Class 11: Binomial Theorem

$\displaystyle \textbf{Question 1: }\text{Find the }11^{\text{th}}\text{ term from the beginning and the }11^{\text{th}}\text{ term from the}$
$\displaystyle \text{end in the expansion of }\left(2x-\frac{1}{x^2}\right)^{25}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(2x-\frac{1}{x^2}\right)^{25}$
$\displaystyle \text{The number of terms in the expansion}=25+1=26.$
$\displaystyle \text{Therefore, the }11^{\text{th}}\text{ term from the end is the }(26-11+1)^{\text{th}}\text{ term}$
$\displaystyle \text{from the beginning, i.e. the }16^{\text{th}}\text{ term from the beginning.}$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{n-r}b^r.$
$\displaystyle \text{The }11^{\text{th}}\text{ term from the beginning is }T_{11}.$
$\displaystyle T_{11}={} ^{25}C_{10}(2x)^{25-10}\left(-\frac{1}{x^2}\right)^{10}$
$\displaystyle ={}^{25}C_{10}(2x)^{15}\left(\frac{1}{x^{20}}\right)$
$\displaystyle ={}^{25}C_{10}\frac{2^{15}}{x^5}$
$\displaystyle \text{The }11^{\text{th}}\text{ term from the end is }T_{16}.$
$\displaystyle T_{16}={} ^{25}C_{15}(2x)^{25-15}\left(-\frac{1}{x^2}\right)^{15}$
$\displaystyle ={}^{25}C_{15}(2x)^{10}\left(-\frac{1}{x^{30}}\right)$
$\displaystyle =-{}^{25}C_{15}\frac{2^{10}}{x^{20}}$
$\displaystyle \therefore \text{The }11^{\text{th}}\text{ term from the beginning is }{}^{25}C_{10}\frac{2^{15}}{x^5}.$
$\displaystyle \text{The }11^{\text{th}}\text{ term from the end is }-{}^{25}C_{15}\frac{2^{10}}{x^{20}}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the }7^{\text{th}}\text{ term in the expansion of }\left(3x^2-\frac{1}{x^3}\right)^{10}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(3x^2-\frac{1}{x^3}\right)^{10}$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{\,n-r}b^r.$
$\displaystyle T_7=T_{6+1}={} ^{10}C_6(3x^2)^{10-6}\left(-\frac{1}{x^3}\right)^6$
$\displaystyle ={}^{10}C_6\left(\frac{3^4x^8}{x^{18}}\right)$
$\displaystyle ={}^{10}C_6\left(\frac{3^4}{x^{10}}\right)$
$\displaystyle =210\times\frac{81}{x^{10}}$
$\displaystyle =\frac{17010}{x^{10}}$
$\displaystyle \therefore \text{The }7^{\text{th}}\text{ term is }\frac{17010}{x^{10}}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the }5^{\text{th}}\text{ term from the end in the expansion of }\left(3x-\frac{1}{x^2}\right)^{10}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(3x-\frac{1}{x^2}\right)^{10}$
$\displaystyle \text{The number of terms in the expansion}=10+1=11.$
$\displaystyle \text{Hence, the }5^{\text{th}}\text{ term from the end is the }(11-5+1)^{\text{th}}\text{ term}$
$\displaystyle \text{from the beginning, i.e. the }7^{\text{th}}\text{ term.}$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{\,n-r}b^r.$
$\displaystyle T_7=T_{6+1}={} ^{10}C_6(3x)^{10-6}\left(-\frac{1}{x^2}\right)^6$
$\displaystyle ={}^{10}C_6(3^4x^4)\left(\frac{1}{x^{12}}\right)$
$\displaystyle =210\times\frac{81}{x^8}$
$\displaystyle =\frac{17010}{x^8}$
$\displaystyle \therefore \text{The }5^{\text{th}}\text{ term from the end is }\frac{17010}{x^8}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the }8^{\text{th}}\text{ term in the expansion of }\left(x^{\frac{3}{2}}y^{\frac{1}{2}}-x^{\frac{1}{2}}y^{\frac{3}{2}}\right)^{10}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(x^{\frac{3}{2}}y^{\frac{1}{2}}-x^{\frac{1}{2}}y^{\frac{3}{2}}\right)^{10}$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{\,n-r}b^r.$
$\displaystyle T_8=T_{7+1}={} ^{10}C_7\left(x^{\frac{3}{2}}y^{\frac{1}{2}}\right)^3\left(-x^{\frac{1}{2}}y^{\frac{3}{2}}\right)^7$
$\displaystyle =-{}^{10}C_7x^{\frac{9}{2}}y^{\frac{3}{2}}x^{\frac{7}{2}}y^{\frac{21}{2}}$
$\displaystyle =-120x^8y^{12}$
$\displaystyle \therefore \text{The }8^{\text{th}}\text{ term is }-120x^8y^{12}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the }7^{\text{th}}\text{ term in the expansion of }\left(\frac{4x}{5}+\frac{5}{2x}\right)^8.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(\frac{4x}{5}+\frac{5}{2x}\right)^8$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{\,n-r}b^r.$
$\displaystyle T_7=T_{6+1}={}^8C_6\left(\frac{4x}{5}\right)^2\left(\frac{5}{2x}\right)^6$
$\displaystyle =28\times\frac{4^2x^2}{5^2}\times\frac{5^6}{2^6x^6}$
$\displaystyle =\frac{4375}{x^4}$
$\displaystyle \therefore \text{The }7^{\text{th}}\text{ term is }\frac{4375}{x^4}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Find the }4^{\text{th}}\text{ term from the beginning and the }4^{\text{th}}\text{ term from the}$
$\displaystyle \text{end in the expansion of }\left(x+\frac{2}{x}\right)^9.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(x+\frac{2}{x}\right)^9$
$\displaystyle \text{The number of terms in the expansion}=9+1=10.$
$\displaystyle \text{Therefore, the }4^{\text{th}}\text{ term from the end is the }(10-4+1)^{\text{th}}\text{ term}$
$\displaystyle \text{from the beginning, i.e. the }7^{\text{th}}\text{ term.}$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{\,n-r}b^r.$
$\displaystyle \text{The }4^{\text{th}}\text{ term from the beginning is }T_4.$
$\displaystyle T_4=T_{3+1}={}^9C_3x^{9-3}\left(\frac{2}{x}\right)^3$
$\displaystyle =84x^6\left(\frac{8}{x^3}\right)$
$\displaystyle =672x^3$
$\displaystyle \text{The }4^{\text{th}}\text{ term from the end is }T_7.$
$\displaystyle T_7=T_{6+1}={}^9C_6x^{9-6}\left(\frac{2}{x}\right)^6$
$\displaystyle =84x^3\left(\frac{64}{x^6}\right)$
$\displaystyle =\frac{5376}{x^3}$
$\displaystyle \therefore \text{The }4^{\text{th}}\text{ term from the beginning is }672x^3.$
$\displaystyle \text{The }4^{\text{th}}\text{ term from the end is }\frac{5376}{x^3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the }4^{\text{th}}\text{ term from the end in the expansion of}$
$\displaystyle \left(\frac{4x}{5}-\frac{5}{2x}\right)^9.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(\frac{4x}{5}-\frac{5}{2x}\right)^9$
$\displaystyle \text{The number of terms in the expansion}=9+1=10.$
$\displaystyle \text{Therefore, the }4^{\text{th}}\text{ term from the end is the }(10-4+1)^{\text{th}}\text{ term}$
$\displaystyle \text{from the beginning, i.e. the }7^{\text{th}}\text{ term.}$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{\,n-r}b^r.$
$\displaystyle T_7=T_{6+1}={}^9C_6\left(\frac{4x}{5}\right)^3\left(-\frac{5}{2x}\right)^6$
$\displaystyle =84\left(\frac{4^3x^3}{5^3}\right)\left(\frac{5^6}{2^6x^6}\right)$
$\displaystyle =84\left(\frac{125}{x^3}\right)$
$\displaystyle =\frac{10500}{x^3}$
$\displaystyle \therefore \text{The }4^{\text{th}}\text{ term from the end is }\frac{10500}{x^3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find the }7^{\text{th}}\text{ term from the end in the expansion of}$
$\displaystyle \left(2x^2-\frac{3}{2x}\right)^8.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(2x^2-\frac{3}{2x}\right)^8$
$\displaystyle \text{The number of terms in the expansion}=8+1=9.$
$\displaystyle \text{Therefore, the }7^{\text{th}}\text{ term from the end is the }(9-7+1)^{\text{th}}\text{ term}$
$\displaystyle \text{from the beginning, i.e. the }3^{\text{rd}}\text{ term.}$
$\displaystyle \text{For }(a+b)^n,\quad T_{r+1}={}^nC_r a^{\,n-r}b^r.$
$\displaystyle T_3=T_{2+1}={}^8C_2(2x^2)^{8-2}\left(-\frac{3}{2x}\right)^2$
$\displaystyle =28\cdot2^6x^{12}\cdot\frac{3^2}{2^2x^2}$
$\displaystyle =28\cdot16\cdot9\,x^{10}$
$\displaystyle =4032x^{10}$
$\displaystyle \therefore \text{The }7^{\text{th}}\text{ term from the end is }4032x^{10}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the coefficient of:}$
$\displaystyle \text{i) }x^{10}\text{ in the expansion of }\left(2x^2-\frac{1}{x}\right)^{20}$
$\displaystyle \text{ii) }x^7\text{ in the expansion of }\left(x-\frac{1}{x^2}\right)^{40}$
$\displaystyle \text{iii) }x^{-15}\text{ in the expansion of }\left(3x^2-\frac{a}{3x^3}\right)^{10}$
$\displaystyle \text{iv) }x^9\text{ in the expansion of }\left(x^2-\frac{1}{3x}\right)^9$
$\displaystyle \text{v) }x^m\text{ in the expansion of }\left(x+\frac{1}{x}\right)^n$
$\displaystyle \text{vi) }x\text{ in the expansion of }(1-2x^3+3x^5)\left(1+\frac{1}{x}\right)^8$
$\displaystyle \text{vii) }a^5b^7\text{ in the expansion of }(a-2b)^{12}$
$\displaystyle \text{viii) }x\text{ in the expansion of }(1-3x+7x^2)(1-x)^{16}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }\text{The general term in }\left(2x^2-\frac{1}{x}\right)^{20}\text{ is}$
$\displaystyle T_{r+1}={} ^{20}C_r(2x^2)^{20-r}\left(-\frac{1}{x}\right)^r$
$\displaystyle =(-1)^r{}^{20}C_r2^{20-r}x^{40-3r}$
$\displaystyle \text{For the term containing }x^{10},$
$\displaystyle 40-3r=10$
$\displaystyle \Rightarrow r=10$
$\displaystyle \therefore \text{Coefficient of }x^{10}=(-1)^{10}{}^{20}C_{10}2^{10}$
$\displaystyle ={}^{20}C_{10}2^{10}$
$\displaystyle \\$

$\displaystyle \text{ii) }\text{The general term in }\left(x-\frac{1}{x^2}\right)^{40}\text{ is}$
$\displaystyle T_{r+1}={} ^{40}C_rx^{40-r}\left(-\frac{1}{x^2}\right)^r$
$\displaystyle =(-1)^r{}^{40}C_rx^{40-3r}$
$\displaystyle \text{For the term containing }x^7,$
$\displaystyle 40-3r=7$
$\displaystyle \Rightarrow r=11$
$\displaystyle \therefore \text{Coefficient of }x^7=(-1)^{11}{}^{40}C_{11}$
$\displaystyle =-{}^{40}C_{11}$
$\displaystyle \\$

$\displaystyle \text{iii) }\text{The general term in }\left(3x^2-\frac{a}{3x^3}\right)^{10}\text{ is}$
$\displaystyle T_{r+1}={} ^{10}C_r(3x^2)^{10-r}\left(-\frac{a}{3x^3}\right)^r$
$\displaystyle =(-1)^r{}^{10}C_r3^{10-2r}a^rx^{20-5r}$
$\displaystyle \text{For the term containing }x^{-15},$
$\displaystyle 20-5r=-15$
$\displaystyle \Rightarrow r=7$
$\displaystyle \therefore \text{Coefficient of }x^{-15}=(-1)^7{}^{10}C_73^{-4}a^7$
$\displaystyle =-\frac{{}^{10}C_7a^7}{3^4}$
$\displaystyle =-\frac{120a^7}{81}$
$\displaystyle =-\frac{40a^7}{27}$
$\displaystyle \\$

$\displaystyle \text{iv) }\text{The general term in }\left(x^2-\frac{1}{3x}\right)^9\text{ is}$
$\displaystyle T_{r+1}={} ^9C_r(x^2)^{9-r}\left(-\frac{1}{3x}\right)^r$
$\displaystyle =(-1)^r{}^9C_r\frac{x^{18-3r}}{3^r}$
$\displaystyle \text{For the term containing }x^9,$
$\displaystyle 18-3r=9$
$\displaystyle \Rightarrow r=3$
$\displaystyle \therefore \text{Coefficient of }x^9=(-1)^3{}^9C_3\frac{1}{3^3}$
$\displaystyle =-\frac{84}{27}$
$\displaystyle =-\frac{28}{9}$
$\displaystyle \\$

$\displaystyle \text{v) }\text{The general term in }\left(x+\frac{1}{x}\right)^n\text{ is}$
$\displaystyle T_{r+1}={}^nC_rx^{n-r}\left(\frac{1}{x}\right)^r$
$\displaystyle ={}^nC_rx^{n-2r}$
$\displaystyle \text{For the term containing }x^m,$
$\displaystyle n-2r=m$
$\displaystyle \Rightarrow r=\frac{n-m}{2}$
$\displaystyle \therefore \text{Coefficient of }x^m={}^nC_{\frac{n-m}{2}}$
$\displaystyle =\frac{n!}{\left(\frac{n-m}{2}\right)!\left(\frac{n+m}{2}\right)!},$
$\displaystyle \text{provided }n-m\text{ is even and }-n\leq m\leq n.$
$\displaystyle \text{Otherwise, the coefficient of }x^m\text{ is }0.$
$\displaystyle \\$

$\displaystyle \text{vi) }\text{Given expression: }(1-2x^3+3x^5)\left(1+\frac{1}{x}\right)^8$
$\displaystyle \text{To obtain }x,\text{ the required products are}$
$\displaystyle -2x^3\left[{}^8C_2\left(\frac{1}{x}\right)^2\right]$
$\displaystyle \text{and }3x^5\left[{}^8C_4\left(\frac{1}{x}\right)^4\right].$
$\displaystyle \text{Therefore, the coefficient of }x$
$\displaystyle =-2{}^8C_2+3{}^8C_4$
$\displaystyle =-2(28)+3(70)$
$\displaystyle =-56+210$
$\displaystyle =154$
$\displaystyle \\$

$\displaystyle \text{vii) }\text{The general term in }(a-2b)^{12}\text{ is}$
$\displaystyle T_{r+1}={} ^{12}C_ra^{12-r}(-2b)^r$
$\displaystyle \text{For the term containing }a^5b^7,$
$\displaystyle 12-r=5$
$\displaystyle \Rightarrow r=7$
$\displaystyle \therefore \text{Coefficient of }a^5b^7={} ^{12}C_7(-2)^7$
$\displaystyle =792(-128)$
$\displaystyle =-101376$
$\displaystyle \\$

$\displaystyle \text{viii) }\text{Given expression: }(1-3x+7x^2)(1-x)^{16}$
$\displaystyle (1-x)^{16}=1-16x+\text{terms containing higher powers of }x$
$\displaystyle \text{The terms containing }x\text{ in the product are }-16x\text{ and }-3x.$
$\displaystyle \therefore \text{Coefficient of }x=-16-3$
$\displaystyle =-19$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Which term in the expansion of }\left\{\left(\frac{x}{\sqrt{y}}\right)^{\frac13}+\left(\frac{y}{x^{\frac13}}\right)^{\frac12}\right\}^{21}$
$\displaystyle \text{contains }x\text{ and }y\text{ raised to one and the same power?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left\{\left(\frac{x}{\sqrt{y}}\right)^{\frac13}+\left(\frac{y}{x^{\frac13}}\right)^{\frac12}\right\}^{21}$
$\displaystyle \text{The general term is}$
$\displaystyle T_{r+1}={} ^{21}C_r\left[\left(\frac{x}{\sqrt{y}}\right)^{\frac13}\right]^{21-r}$
$\displaystyle \qquad\times\left[\left(\frac{y}{x^{\frac13}}\right)^{\frac12}\right]^r$
$\displaystyle ={}^{21}C_r\left(\frac{x^{\frac{21-r}{3}}}{y^{\frac{21-r}{6}}}\right)\left(\frac{y^{\frac r2}}{x^{\frac r6}}\right)$
$\displaystyle ={}^{21}C_r x^{\frac{21-r}{3}-\frac r6}y^{\frac r2-\frac{21-r}{6}}$
$\displaystyle ={}^{21}C_r x^{7-\frac r2}y^{\frac{2r}{3}-\frac72}$
$\displaystyle \text{For }x\text{ and }y\text{ to have the same power,}$
$\displaystyle 7-\frac r2=\frac{2r}{3}-\frac72$
$\displaystyle \Rightarrow 42-3r=4r-21$
$\displaystyle \Rightarrow 7r=63$
$\displaystyle \Rightarrow r=9$
$\displaystyle \therefore T_{r+1}=T_{10}.$
$\displaystyle \therefore \text{The }10^{\text{th}}\text{ term contains }x\text{ and }y\text{ raised to the same power.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Does the expansion of }\left(2x^2-\frac{1}{x}\right)^{20}\text{ contain any term involving }x^9?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(2x^2-\frac{1}{x}\right)^{20}$
$\displaystyle \text{Suppose }x^9\text{ occurs in the }(r+1)^{\text{th}}\text{ term of the expansion.}$
$\displaystyle T_{r+1}={} ^{20}C_r(2x^2)^{20-r}\left(-\frac{1}{x}\right)^r$
$\displaystyle =(-1)^r{}^{20}C_r2^{20-r}x^{40-2r-r}$
$\displaystyle =(-1)^r{}^{20}C_r2^{20-r}x^{40-3r}$
$\displaystyle \text{For the term to contain }x^9,$
$\displaystyle 40-3r=9$
$\displaystyle \Rightarrow r=\frac{31}{3}$
$\displaystyle \text{Since }r\text{ must be an integer, this is not possible.}$
$\displaystyle \therefore \text{The expansion does not contain any term involving }x^9.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Show that the expansion of }\left(x^2+\frac{1}{x}\right)^{12}\text{ does not contain any}$
$\displaystyle \text{term involving }x^{-1}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(x^2+\frac{1}{x}\right)^{12}$
$\displaystyle \text{Suppose }x^{-1}\text{ occurs in the }(r+1)^{\text{th}}\text{ term of the expansion.}$
$\displaystyle T_{r+1}={} ^{12}C_r(x^2)^{12-r}\left(\frac{1}{x}\right)^r$
$\displaystyle ={}^{12}C_rx^{24-2r-r}$
$\displaystyle ={}^{12}C_rx^{24-3r}$
$\displaystyle \text{For the term to contain }x^{-1},$
$\displaystyle 24-3r=-1$
$\displaystyle \Rightarrow r=\frac{25}{3}$
$\displaystyle \text{Since }r\text{ must be an integer, this is not possible.}$
$\displaystyle \therefore \text{The expansion does not contain any term involving }x^{-1}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Find the middle term in the expansion of:}$
$\displaystyle \text{i) }\left(\frac{2x}{3}-\frac{3}{2x}\right)^{20}\qquad\text{ii) }\left(\frac{a}{x}+bx\right)^{12}$
$\displaystyle \text{iii) }\left(x^2-\frac{2}{x}\right)^{10}\qquad\text{iv) }\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }\left(\frac{2x}{3}-\frac{3}{2x}\right)^{20}$
$\displaystyle \text{Here, }n=20\text{ is even. Therefore, the }\left(\frac{20}{2}+1\right)^{\text{th}}$
$\displaystyle \text{term, i.e. the }11^{\text{th}}\text{ term, is the middle term.}$
$\displaystyle T_{11}=T_{10+1}={} ^{20}C_{10}\left(\frac{2x}{3}\right)^{10}\left(-\frac{3}{2x}\right)^{10}$
$\displaystyle ={}^{20}C_{10}\left(\frac{2}{3}\right)^{10}x^{10}\left(\frac{3}{2}\right)^{10}\frac{1}{x^{10}}$
$\displaystyle ={}^{20}C_{10}$
$\displaystyle =184756$
$\displaystyle \therefore \text{The middle term is }184756.$
$\displaystyle \\$

$\displaystyle \text{ii) }\left(\frac{a}{x}+bx\right)^{12}$
$\displaystyle \text{Here, }n=12\text{ is even. Therefore, the }\left(\frac{12}{2}+1\right)^{\text{th}}$
$\displaystyle \text{term, i.e. the }7^{\text{th}}\text{ term, is the middle term.}$
$\displaystyle T_7=T_{6+1}={} ^{12}C_6\left(\frac{a}{x}\right)^6(bx)^6$
$\displaystyle ={}^{12}C_6a^6b^6$
$\displaystyle =924a^6b^6$
$\displaystyle \therefore \text{The middle term is }924a^6b^6.$
$\displaystyle \\$

$\displaystyle \text{iii) }\left(x^2-\frac{2}{x}\right)^{10}$
$\displaystyle \text{Here, }n=10\text{ is even. Therefore, the }\left(\frac{10}{2}+1\right)^{\text{th}}$
$\displaystyle \text{term, i.e. the }6^{\text{th}}\text{ term, is the middle term.}$
$\displaystyle T_6=T_{5+1}={} ^{10}C_5(x^2)^5\left(-\frac{2}{x}\right)^5$
$\displaystyle =-{}^{10}C_5x^{10}\frac{2^5}{x^5}$
$\displaystyle =-252(32)x^5$
$\displaystyle =-8064x^5$
$\displaystyle \therefore \text{The middle term is }-8064x^5.$
$\displaystyle \\$

$\displaystyle \text{iv) }\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$
$\displaystyle \text{Here, }n=10\text{ is even. Therefore, the }\left(\frac{10}{2}+1\right)^{\text{th}}$
$\displaystyle \text{term, i.e. the }6^{\text{th}}\text{ term, is the middle term.}$
$\displaystyle T_6=T_{5+1}={} ^{10}C_5\left(\frac{x}{a}\right)^5\left(-\frac{a}{x}\right)^5$
$\displaystyle =-{}^{10}C_5\left(\frac{x^5}{a^5}\right)\left(\frac{a^5}{x^5}\right)$
$\displaystyle =-{}^{10}C_5$
$\displaystyle =-252$
$\displaystyle \therefore \text{The middle term is }-252.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Find the middle terms in the expansion of:}$
$\displaystyle \text{i) }\left(3x-\frac{x^3}{6}\right)^9\qquad\text{ii) }\left(2x^2-\frac{1}{x}\right)^7$
$\displaystyle \text{iii) }\left(3x-\frac{2}{x^2}\right)^{15}\qquad\text{iv) }\left(x^4-\frac{1}{x^3}\right)^{11}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }\text{Given expression: }\left(3x-\frac{x^3}{6}\right)^9$
$\displaystyle \text{Here, }n=9\text{ is odd. Therefore, the }\left(\frac{9+1}{2}\right)^{\text{th}}\text{ and}$
$\displaystyle \left(\frac{9+1}{2}+1\right)^{\text{th}}\text{ terms, i.e. the }5^{\text{th}}\text{ and }6^{\text{th}}\text{ terms, are the middle terms.}$
$\displaystyle T_5=T_{4+1}={}^9C_4(3x)^{9-4}\left(-\frac{x^3}{6}\right)^4$
$\displaystyle ={}^9C_4(3x)^5\frac{x^{12}}{6^4}$
$\displaystyle =126\cdot\frac{3^5}{6^4}x^{17}$
$\displaystyle =\frac{189}{8}x^{17}$
$\displaystyle T_6=T_{5+1}={}^9C_5(3x)^{9-5}\left(-\frac{x^3}{6}\right)^5$
$\displaystyle =-{}^9C_5(3x)^4\frac{x^{15}}{6^5}$
$\displaystyle =-126\cdot\frac{3^4}{6^5}x^{19}$
$\displaystyle =-\frac{21}{16}x^{19}$
$\displaystyle \therefore \text{The middle terms are }\frac{189}{8}x^{17}\text{ and }-\frac{21}{16}x^{19}.$
$\displaystyle \\$

$\displaystyle \text{ii) }\text{Given expression: }\left(2x^2-\frac{1}{x}\right)^7$
$\displaystyle \text{Here, }n=7\text{ is odd. Therefore, the }\left(\frac{7+1}{2}\right)^{\text{th}}\text{ and}$
$\displaystyle \left(\frac{7+1}{2}+1\right)^{\text{th}}\text{ terms, i.e. the }4^{\text{th}}\text{ and }5^{\text{th}}\text{ terms, are the middle terms.}$
$\displaystyle T_4=T_{3+1}={}^7C_3(2x^2)^{7-3}\left(-\frac{1}{x}\right)^3$
$\displaystyle =-{}^7C_3(2x^2)^4\frac{1}{x^3}$
$\displaystyle =-35\cdot16x^5$
$\displaystyle =-560x^5$
$\displaystyle T_5=T_{4+1}={}^7C_4(2x^2)^{7-4}\left(-\frac{1}{x}\right)^4$
$\displaystyle ={}^7C_4(2x^2)^3\frac{1}{x^4}$
$\displaystyle =35\cdot8x^2$
$\displaystyle =280x^2$
$\displaystyle \therefore \text{The middle terms are }-560x^5\text{ and }280x^2.$
$\displaystyle \\$

$\displaystyle \text{iii) }\text{Given expression: }\left(3x-\frac{2}{x^2}\right)^{15}$
$\displaystyle \text{Here, }n=15\text{ is odd. Therefore, the }\left(\frac{15+1}{2}\right)^{\text{th}}\text{ and}$
$\displaystyle \left(\frac{15+1}{2}+1\right)^{\text{th}}\text{ terms, i.e. the }8^{\text{th}}\text{ and }9^{\text{th}}\text{ terms, are the middle terms.}$
$\displaystyle T_8=T_{7+1}={} ^{15}C_7(3x)^{15-7}\left(-\frac{2}{x^2}\right)^7$
$\displaystyle =-{}^{15}C_7\cdot3^8x^8\cdot\frac{2^7}{x^{14}}$
$\displaystyle =-\frac{6435\cdot3^8\cdot2^7}{x^6}$
$\displaystyle T_9=T_{8+1}={} ^{15}C_8(3x)^{15-8}\left(-\frac{2}{x^2}\right)^8$
$\displaystyle ={}^{15}C_8\cdot3^7x^7\cdot\frac{2^8}{x^{16}}$
$\displaystyle =\frac{6435\cdot3^7\cdot2^8}{x^9}$
$\displaystyle \therefore \text{The middle terms are }-\frac{6435\cdot3^8\cdot2^7}{x^6}\text{ and}$
$\displaystyle \frac{6435\cdot3^7\cdot2^8}{x^9}.$
$\displaystyle \\$

$\displaystyle \text{iv) }\text{Given expression: }\left(x^4-\frac{1}{x^3}\right)^{11}$
$\displaystyle \text{Here, }n=11\text{ is odd. Therefore, the }\left(\frac{11+1}{2}\right)^{\text{th}}\text{ and}$
$\displaystyle \left(\frac{11+1}{2}+1\right)^{\text{th}}\text{ terms, i.e. the }6^{\text{th}}\text{ and }7^{\text{th}}\text{ terms, are the middle terms.}$
$\displaystyle T_6=T_{5+1}={} ^{11}C_5(x^4)^{11-5}\left(-\frac{1}{x^3}\right)^5$
$\displaystyle =-{}^{11}C_5x^{24}\frac{1}{x^{15}}$
$\displaystyle =-462x^9$
$\displaystyle T_7=T_{6+1}={} ^{11}C_6(x^4)^{11-6}\left(-\frac{1}{x^3}\right)^6$
$\displaystyle ={}^{11}C_6x^{20}\frac{1}{x^{18}}$
$\displaystyle =462x^2$
$\displaystyle \therefore \text{The middle terms are }-462x^9\text{ and }462x^2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Find the middle term or terms in the expansion of:}$
$\displaystyle \text{i) }\left(x-\frac{1}{x}\right)^{10}\qquad\text{ii) }(1-2x+x^2)^n$
$\displaystyle \text{iii) }(1+3x+3x^2+x^3)^{2n}\qquad\text{iv) }\left(2x-\frac{x^2}{4}\right)^9$
$\displaystyle \text{v) }\left(x-\frac{1}{x}\right)^{2n+1}\qquad\text{vi) }\left(\frac{x}{3}+9y\right)^{10}$
$\displaystyle \text{vii) }\left(3-\frac{x^3}{6}\right)^7\qquad\text{viii) }\left(2ax-\frac{b}{x^2}\right)^{12}$
$\displaystyle \text{ix) }\left(\frac{p}{x}+\frac{x}{p}\right)^9\qquad\text{x) }\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }\text{Given expression: }\left(x-\frac{1}{x}\right)^{10}$
$\displaystyle \text{Since the exponent }10\text{ is even, the }6^{\text{th}}\text{ term is the middle term.}$
$\displaystyle T_6=T_{5+1}={} ^{10}C_5x^{10-5}\left(-\frac{1}{x}\right)^5$
$\displaystyle =-{}^{10}C_5\frac{x^5}{x^5}$
$\displaystyle =-252$
$\displaystyle \therefore \text{The middle term is }-252.$
$\displaystyle \\$

$\displaystyle \text{ii) }\text{Given expression: }(1-2x+x^2)^n$
$\displaystyle =(1-x)^{2n}$
$\displaystyle \text{Since the exponent }2n\text{ is even, the }(n+1)^{\text{th}}\text{ term is the middle term.}$
$\displaystyle T_{n+1}={} ^{2n}C_n(1)^{2n-n}(-x)^n$
$\displaystyle =(-1)^n{}^{2n}C_nx^n$
$\displaystyle =(-1)^n\frac{(2n)!}{(n!)^2}x^n$
$\displaystyle \therefore \text{The middle term is }(-1)^n\frac{(2n)!}{(n!)^2}x^n.$
$\displaystyle \\$

$\displaystyle \text{iii) }\text{Given expression: }(1+3x+3x^2+x^3)^{2n}$
$\displaystyle =\left[(1+x)^3\right]^{2n}$
$\displaystyle =(1+x)^{6n}$
$\displaystyle \text{Since the exponent }6n\text{ is even, the }(3n+1)^{\text{th}}\text{ term is the middle term.}$
$\displaystyle T_{3n+1}={} ^{6n}C_{3n}x^{3n}$
$\displaystyle =\frac{(6n)!}{[(3n)!]^2}x^{3n}$
$\displaystyle \therefore \text{The middle term is }\frac{(6n)!}{[(3n)!]^2}x^{3n}.$
$\displaystyle \\$

$\displaystyle \text{iv) }\text{Given expression: }\left(2x-\frac{x^2}{4}\right)^9$
$\displaystyle \text{Since the exponent }9\text{ is odd, the }5^{\text{th}}\text{ and }6^{\text{th}}\text{ terms are the middle terms.}$
$\displaystyle T_5={}^9C_4(2x)^5\left(-\frac{x^2}{4}\right)^4$
$\displaystyle =126\cdot\frac{2^5}{4^4}x^{13}$
$\displaystyle =\frac{63}{4}x^{13}$
$\displaystyle T_6={}^9C_5(2x)^4\left(-\frac{x^2}{4}\right)^5$
$\displaystyle =-126\cdot\frac{2^4}{4^5}x^{14}$
$\displaystyle =-\frac{63}{32}x^{14}$
$\displaystyle \therefore \text{The middle terms are }\frac{63}{4}x^{13}\text{ and }-\frac{63}{32}x^{14}.$
$\displaystyle \\$

$\displaystyle \text{v) }\text{Given expression: }\left(x-\frac{1}{x}\right)^{2n+1}$
$\displaystyle \text{Since the exponent }2n+1\text{ is odd, the }(n+1)^{\text{th}}\text{ and }(n+2)^{\text{th}}$
$\displaystyle \text{terms are the middle terms.}$
$\displaystyle T_{n+1}={} ^{2n+1}C_nx^{n+1}\left(-\frac{1}{x}\right)^n$
$\displaystyle =(-1)^n{}^{2n+1}C_nx$
$\displaystyle T_{n+2}={} ^{2n+1}C_{n+1}x^n\left(-\frac{1}{x}\right)^{n+1}$
$\displaystyle =(-1)^{n+1}{}^{2n+1}C_{n+1}\frac{1}{x}$
$\displaystyle =(-1)^{n+1}{}^{2n+1}C_n\frac{1}{x}$
$\displaystyle \therefore \text{The middle terms are }(-1)^n{}^{2n+1}C_nx\text{ and}$
$\displaystyle (-1)^{n+1}{}^{2n+1}C_n\frac{1}{x}.$
$\displaystyle \\$

$\displaystyle \text{vi) }\text{Given expression: }\left(\frac{x}{3}+9y\right)^{10}$
$\displaystyle \text{Since the exponent }10\text{ is even, the }6^{\text{th}}\text{ term is the middle term.}$
$\displaystyle T_6={} ^{10}C_5\left(\frac{x}{3}\right)^5(9y)^5$
$\displaystyle =252\cdot\frac{x^5}{3^5}\cdot9^5y^5$
$\displaystyle =252\cdot3^5x^5y^5$
$\displaystyle =61236x^5y^5$
$\displaystyle \therefore \text{The middle term is }61236x^5y^5.$
$\displaystyle \\$

$\displaystyle \text{vii) }\text{Given expression: }\left(3-\frac{x^3}{6}\right)^7$
$\displaystyle \text{Since the exponent }7\text{ is odd, the }4^{\text{th}}\text{ and }5^{\text{th}}\text{ terms are the middle terms.}$
$\displaystyle T_4={}^7C_3(3)^4\left(-\frac{x^3}{6}\right)^3$
$\displaystyle =-35\cdot3^4\frac{x^9}{6^3}$
$\displaystyle =-\frac{105}{8}x^9$
$\displaystyle T_5={}^7C_4(3)^3\left(-\frac{x^3}{6}\right)^4$
$\displaystyle =35\cdot3^3\frac{x^{12}}{6^4}$
$\displaystyle =\frac{35}{48}x^{12}$
$\displaystyle \therefore \text{The middle terms are }-\frac{105}{8}x^9\text{ and }\frac{35}{48}x^{12}.$
$\displaystyle \\$

$\displaystyle \text{viii) }\text{Given expression: }\left(2ax-\frac{b}{x^2}\right)^{12}$
$\displaystyle \text{Since the exponent }12\text{ is even, the }7^{\text{th}}\text{ term is the middle term.}$
$\displaystyle T_7={} ^{12}C_6(2ax)^6\left(-\frac{b}{x^2}\right)^6$
$\displaystyle =924\cdot2^6a^6x^6\frac{b^6}{x^{12}}$
$\displaystyle =\frac{59136a^6b^6}{x^6}$
$\displaystyle \therefore \text{The middle term is }\frac{59136a^6b^6}{x^6}.$
$\displaystyle \\$

$\displaystyle \text{ix) }\text{Given expression: }\left(\frac{p}{x}+\frac{x}{p}\right)^9$
$\displaystyle \text{Since the exponent }9\text{ is odd, the }5^{\text{th}}\text{ and }6^{\text{th}}\text{ terms are the middle terms.}$
$\displaystyle T_5={}^9C_4\left(\frac{p}{x}\right)^5\left(\frac{x}{p}\right)^4$
$\displaystyle =126\frac{p}{x}$
$\displaystyle T_6={}^9C_5\left(\frac{p}{x}\right)^4\left(\frac{x}{p}\right)^5$
$\displaystyle =126\frac{x}{p}$
$\displaystyle \therefore \text{The middle terms are }\frac{126p}{x}\text{ and }\frac{126x}{p}.$
$\displaystyle \\$

$\displaystyle \text{x) }\text{Given expression: }\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$
$\displaystyle \text{Since the exponent }10\text{ is even, the }6^{\text{th}}\text{ term is the middle term.}$
$\displaystyle T_6={} ^{10}C_5\left(\frac{x}{a}\right)^5\left(-\frac{a}{x}\right)^5$
$\displaystyle =-{}^{10}C_5\left(\frac{x^5}{a^5}\right)\left(\frac{a^5}{x^5}\right)$
$\displaystyle =-{}^{10}C_5$
$\displaystyle =-252$
$\displaystyle \therefore \text{The middle term is }-252.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Find the term independent of }x\text{ in the expansion of the following}$
$\displaystyle \text{expressions:}$
$\displaystyle \text{i) }\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9\qquad \text{ii) }\left(2x+\frac{1}{3x^2}\right)^9$
$\displaystyle \text{iii) }\left(2x^2-\frac{3}{x^3}\right)^{25}\qquad \text{iv) }\left(3x-\frac{2}{x^2}\right)^{15}$
$\displaystyle \text{v) }\left(\sqrt{\frac{x}{3}}+\frac{\sqrt{3}}{2x^2}\right)^{10}\qquad \text{vi) }\left(x-\frac{1}{x^2}\right)^{3n}$
$\displaystyle \text{vii) }\left(\frac{1}{2}x^{\frac{1}{3}}+x^{-\frac{1}{5}}\right)^{8}$
$\displaystyle \text{viii) }(1+x+2x^3)\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9$
$\displaystyle \text{ix) }\left(\sqrt[3]{x}+\frac{1}{2}\sqrt[3]{x}\right)^{18},\ x>2$
$\displaystyle \text{x) }\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^6$
$\displaystyle \text{Answer:}$

$\displaystyle \text{i) Given expression: }\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^9\mathrm{C}_r}\left(\frac{3}{2}x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^r$
$\displaystyle =(-1)^r{^9\mathrm{C}_r}\left(\frac{3}{2}\right)^{9-r}\frac{x^{18-2r}}{3^rx^r}$
$\displaystyle =(-1)^r{^9\mathrm{C}_r}\left(\frac{3}{2}\right)^{9-r}\frac{x^{18-3r}}{3^r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle 18-3r=0$
$\displaystyle \Rightarrow r=6$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }7^{\text{th}}\text{ term.}$
$\displaystyle T_7=(-1)^6{^9\mathrm{C}_6}\left(\frac{3}{2}\right)^3\frac{1}{3^6}$
$\displaystyle ={^9\mathrm{C}_6}\frac{3^3}{2^3\cdot3^6}$
$\displaystyle =84\times\frac{1}{216}$
$\displaystyle =\frac{7}{18}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }\frac{7}{18}.$
$\displaystyle \\$

$\displaystyle \text{ii) Given expression: }\left(2x+\frac{1}{3x^2}\right)^9$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^9\mathrm{C}_r}(2x)^{9-r}\left(\frac{1}{3x^2}\right)^r$
$\displaystyle ={^9\mathrm{C}_r}2^{9-r}\frac{x^{9-r}}{3^rx^{2r}}$
$\displaystyle ={^9\mathrm{C}_r}\frac{2^{9-r}}{3^r}x^{9-3r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle 9-3r=0$
$\displaystyle \Rightarrow r=3$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }4^{\text{th}}\text{ term.}$
$\displaystyle T_4={^9\mathrm{C}_3}\frac{2^6}{3^3}$
$\displaystyle =84\times\frac{64}{27}$
$\displaystyle =\frac{1792}{9}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }\frac{1792}{9}.$
$\displaystyle \\$

$\displaystyle \text{iii) Given expression: }\left(2x^2-\frac{3}{x^3}\right)^{25}$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^{25}\mathrm{C}_r}(2x^2)^{25-r}\left(-\frac{3}{x^3}\right)^r$
$\displaystyle =(-3)^r{^{25}\mathrm{C}_r}2^{25-r}\frac{x^{50-2r}}{x^{3r}}$
$\displaystyle =(-3)^r{^{25}\mathrm{C}_r}2^{25-r}x^{50-5r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle 50-5r=0$
$\displaystyle \Rightarrow r=10$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }11^{\text{th}}\text{ term.}$
$\displaystyle T_{11}=(-3)^{10}{^{25}\mathrm{C}_{10}}2^{25-10}$
$\displaystyle =3^{10}\cdot2^{15}\cdot{^{25}\mathrm{C}_{10}}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }3^{10}\cdot2^{15}\cdot{^{25}\mathrm{C}_{10}}.$
$\displaystyle \\$

$\displaystyle \text{iv) Given expression: }\left(3x-\frac{2}{x^2}\right)^{15}$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^{15}\mathrm{C}_r}(3x)^{15-r}\left(-\frac{2}{x^2}\right)^r$
$\displaystyle =(-1)^r{^{15}\mathrm{C}_r}3^{15-r}2^r\frac{x^{15-r}}{x^{2r}}$
$\displaystyle =(-1)^r{^{15}\mathrm{C}_r}3^{15-r}2^rx^{15-3r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle 15-3r=0$
$\displaystyle \Rightarrow r=5$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }6^{\text{th}}\text{ term.}$
$\displaystyle T_6=(-1)^5{^{15}\mathrm{C}_5}3^{15-5}2^5$
$\displaystyle =-{^{15}\mathrm{C}_5}3^{10}\cdot2^5$
$\displaystyle =-3003\cdot3^{10}\cdot2^5$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }-3003\cdot3^{10}\cdot2^5.$
$\displaystyle \\$

$\displaystyle \text{v) Given expression: }\left(\sqrt{\frac{x}{3}}+\frac{\sqrt{3}}{2x^2}\right)^{10}$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^{10}\mathrm{C}_r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\frac{\sqrt{3}}{2x^2}\right)^r$
$\displaystyle ={^{10}\mathrm{C}_r}\frac{x^{\frac{10-r}{2}}}{3^{\frac{10-r}{2}}}\cdot\frac{3^{\frac{r}{2}}}{2^rx^{2r}}$
$\displaystyle ={^{10}\mathrm{C}_r}\frac{3^{\frac{r}{2}}}{2^r3^{\frac{10-r}{2}}}x^{\frac{10-r}{2}-2r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle \frac{10-r}{2}-2r=0$
$\displaystyle \Rightarrow 10-r-4r=0$
$\displaystyle \Rightarrow r=2$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }3^{\text{rd}}\text{ term.}$
$\displaystyle T_3={^{10}\mathrm{C}_2}\frac{3^{\frac{2}{2}}}{2^2\cdot3^{\frac{10-2}{2}}}$
$\displaystyle ={^{10}\mathrm{C}_2}\frac{3}{4\cdot3^4}$
$\displaystyle =45\times\frac{1}{108}$
$\displaystyle =\frac{5}{12}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }\frac{5}{12}.$
$\displaystyle \\$

$\displaystyle \text{vi) Given expression: }\left(x-\frac{1}{x^2}\right)^{3n}$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^{3n}\mathrm{C}_r}x^{3n-r}\left(-\frac{1}{x^2}\right)^r$
$\displaystyle =(-1)^r{^{3n}\mathrm{C}_r}\frac{x^{3n-r}}{x^{2r}}$
$\displaystyle =(-1)^r{^{3n}\mathrm{C}_r}x^{3n-3r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle 3n-3r=0$
$\displaystyle \Rightarrow r=n$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }(n+1)^{\text{th}}\text{ term.}$
$\displaystyle T_{n+1}=(-1)^n{^{3n}\mathrm{C}_n}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }(-1)^n{^{3n}\mathrm{C}_n}.$
$\displaystyle \\$

$\displaystyle \text{vii) Given expression: }\left(\frac{1}{2}x^{\frac{1}{3}}+x^{-\frac{1}{5}}\right)^8$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^8\mathrm{C}_r}\left(\frac{1}{2}x^{\frac{1}{3}}\right)^{8-r}\left(x^{-\frac{1}{5}}\right)^r$
$\displaystyle ={^8\mathrm{C}_r}\frac{x^{\frac{8-r}{3}}}{2^{8-r}}x^{-\frac{r}{5}}$
$\displaystyle ={^8\mathrm{C}_r}\frac{1}{2^{8-r}}x^{\frac{8-r}{3}-\frac{r}{5}}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle \frac{8-r}{3}-\frac{r}{5}=0$
$\displaystyle \Rightarrow 5(8-r)-3r=0$
$\displaystyle \Rightarrow 40-8r=0$
$\displaystyle \Rightarrow r=5$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }6^{\text{th}}\text{ term.}$
$\displaystyle T_6={^8\mathrm{C}_5}\frac{1}{2^{8-5}}$
$\displaystyle ={^8\mathrm{C}_5}\frac{1}{8}$
$\displaystyle =56\times\frac{1}{8}$
$\displaystyle =7$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }7.$
$\displaystyle \\$

$\displaystyle \text{viii) Given expression: }(1+x+2x^3)\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9$
$\displaystyle \text{The general term in }\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^9\text{ is}$
$\displaystyle T_{r+1}={^9\mathrm{C}_r}\left(\frac{3}{2}x^2\right)^{9-r}\left(-\frac{1}{3x}\right)^r$
$\displaystyle =(-1)^r{^9\mathrm{C}_r}\frac{\left(\frac{3}{2}\right)^{9-r}}{3^r}x^{18-3r}$
$\displaystyle \text{The term independent of }x\text{ is obtained from the terms containing }x^0,x^{-1}\text{ and }x^{-3}.$
$\displaystyle \text{For the term containing }x^0,$
$\displaystyle 18-3r=0$
$\displaystyle \Rightarrow r=6$
$\displaystyle T_7=(-1)^6{^9\mathrm{C}_6}\frac{\left(\frac{3}{2}\right)^3}{3^6}$
$\displaystyle =84\times\frac{27}{8\times729}$
$\displaystyle =\frac{7}{18}$
$\displaystyle \text{For the term containing }x^{-1},$
$\displaystyle 18-3r=-1$
$\displaystyle \Rightarrow r=\frac{19}{3}$
$\displaystyle \text{Since }r\text{ is not an integer, there is no term containing }x^{-1}.$
$\displaystyle \text{For the term containing }x^{-3},$
$\displaystyle 18-3r=-3$
$\displaystyle \Rightarrow r=7$
$\displaystyle T_8=(-1)^7{^9\mathrm{C}_7}\frac{\left(\frac{3}{2}\right)^2}{3^7}x^{-3}$
$\displaystyle =-36\times\frac{9}{4\times2187}x^{-3}$
$\displaystyle =-\frac{1}{27}x^{-3}$
$\displaystyle 2x^3T_8=2x^3\left(-\frac{1}{27}x^{-3}\right)=-\frac{2}{27}$
$\displaystyle \therefore \text{Term independent of }x=\frac{7}{18}-\frac{2}{27}$
$\displaystyle =\frac{21-4}{54}$
$\displaystyle =\frac{17}{54}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }\frac{17}{54}.$
$\displaystyle \\$

$\displaystyle \text{ix) Given expression: }\left(\sqrt[3]{x}+\frac{1}{2\sqrt[3]{x}}\right)^{18},\ x>2$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^{18}\mathrm{C}_r}\left(x^{\frac{1}{3}}\right)^{18-r}\left(\frac{1}{2x^{\frac{1}{3}}}\right)^r$
$\displaystyle ={^{18}\mathrm{C}_r}\frac{x^{\frac{18-r}{3}}}{2^rx^{\frac{r}{3}}}$
$\displaystyle ={^{18}\mathrm{C}_r}\frac{1}{2^r}x^{\frac{18-2r}{3}}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle \frac{18-2r}{3}=0$
$\displaystyle \Rightarrow r=9$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }10^{\text{th}}\text{ term.}$
$\displaystyle T_{10}={^{18}\mathrm{C}_9}\frac{1}{2^9}$
$\displaystyle =\frac{48620}{512}$
$\displaystyle =\frac{12155}{128}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }\frac{12155}{128}.$
$\displaystyle \\$

$\displaystyle \text{x) Given expression: }\left(\frac{3}{2}x^2-\frac{1}{3x}\right)^6$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^6\mathrm{C}_r}\left(\frac{3}{2}x^2\right)^{6-r}\left(-\frac{1}{3x}\right)^r$
$\displaystyle =(-1)^r{^6\mathrm{C}_r}\left(\frac{3}{2}\right)^{6-r}\frac{x^{12-2r}}{3^rx^r}$
$\displaystyle =(-1)^r{^6\mathrm{C}_r}\left(\frac{3}{2}\right)^{6-r}\frac{x^{12-3r}}{3^r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle 12-3r=0$
$\displaystyle \Rightarrow r=4$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }5^{\text{th}}\text{ term.}$
$\displaystyle T_5=(-1)^4{^6\mathrm{C}_4}\left(\frac{3}{2}\right)^2\frac{1}{3^4}$
$\displaystyle =15\times\frac{9}{4}\times\frac{1}{81}$
$\displaystyle =\frac{5}{12}$
$\displaystyle \therefore \text{The required term independent of }x\text{ is }\frac{5}{12}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If the coefficients of the }(2r+4)^{\text{th}}\text{ and }(r-2)^{\text{th}}\text{ terms}$
$\displaystyle \text{in the expansion of }(1+x)^{18}\text{ are equal, find }r.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The general term in the expansion of }(1+x)^{18}\text{ is}$
$\displaystyle T_{k+1}={^{18}\mathrm{C}_k}x^k.$
$\displaystyle \therefore \text{Coefficient of the }(2r+4)^{\text{th}}\text{ term}={^{18}\mathrm{C}_{2r+3}}$
$\displaystyle \text{Coefficient of the }(r-2)^{\text{th}}\text{ term}={^{18}\mathrm{C}_{r-3}}$
$\displaystyle \text{Given, }{^{18}\mathrm{C}_{2r+3}}={^{18}\mathrm{C}_{r-3}}$
$\displaystyle \text{Since }{^n\mathrm{C}_p}={^n\mathrm{C}_q},\text{ either }p=q\text{ or }p+q=n.$
$\displaystyle \text{Case i: }2r+3=r-3$
$\displaystyle \Rightarrow r=-6$
$\displaystyle \text{This is not admissible, since the term numbers must be positive.}$
$\displaystyle \text{Case ii: }(2r+3)+(r-3)=18$
$\displaystyle \Rightarrow 3r=18$
$\displaystyle \Rightarrow r=6$
$\displaystyle \therefore r=6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{If the coefficients of the }(2r+1)^{\text{th}}\text{ and }(r+2)^{\text{th}}\text{ terms}$
$\displaystyle \text{in the expansion of }(1+x)^{43}\text{ are equal, find }r.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The general term in the expansion of }(1+x)^{43}\text{ is}$
$\displaystyle T_{k+1}={^{43}\mathrm{C}_k}x^k.$
$\displaystyle \therefore \text{Coefficient of the }(2r+1)^{\text{th}}\text{ term}={^{43}\mathrm{C}_{2r}}$
$\displaystyle \text{Coefficient of the }(r+2)^{\text{th}}\text{ term}={^{43}\mathrm{C}_{r+1}}$
$\displaystyle \text{Given, }{^{43}\mathrm{C}_{2r}}={^{43}\mathrm{C}_{r+1}}$
$\displaystyle \text{Since }{^n\mathrm{C}_p}={^n\mathrm{C}_q},\text{ either }p=q\text{ or }p+q=n.$
$\displaystyle \text{Case i: }2r=r+1$
$\displaystyle \Rightarrow r=1$
$\displaystyle \text{In this case, both expressions refer to the same term.}$
$\displaystyle \text{Case ii: }2r+(r+1)=43$
$\displaystyle \Rightarrow 3r=42$
$\displaystyle \Rightarrow r=14$
$\displaystyle \therefore r=1\text{ or }r=14.$
$\displaystyle \text{If the two terms are required to be distinct, then }r=14.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Prove that the coefficient of the }(r+1)^{\text{th}}\text{ term in the expansion}$
$\displaystyle \text{of }(1+x)^{n+1}\text{ is equal to the sum of the coefficients of the }r^{\text{th}}\text{ and}$
$\displaystyle \text{the }(r+1)^{\text{th}}\text{ terms in the expansion of }(1+x)^n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In the expansion of }(1+x)^{n+1},$
$\displaystyle T_{r+1}={^{n+1}\mathrm{C}_r}x^r.$
$\displaystyle \therefore \text{Coefficient of the }(r+1)^{\text{th}}\text{ term}={^{n+1}\mathrm{C}_r}.$
$\displaystyle \text{In the expansion of }(1+x)^n,$
$\displaystyle T_r={^n\mathrm{C}_{r-1}}x^{r-1}$
$\displaystyle \text{and }T_{r+1}={^n\mathrm{C}_r}x^r.$
$\displaystyle \therefore \text{Sum of their coefficients}={^n\mathrm{C}_{r-1}}+{^n\mathrm{C}_r}$
$\displaystyle ={^{n+1}\mathrm{C}_r}\qquad\text{[By Pascal's identity]}$
$\displaystyle \therefore \text{The coefficient of the }(r+1)^{\text{th}}\text{ term in }(1+x)^{n+1}$
$\displaystyle \text{equals the sum of the coefficients of the }r^{\text{th}}\text{ and }(r+1)^{\text{th}}\text{ terms in }(1+x)^n.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

Question 20: Prove that the term independent of $\displaystyle x$ in the expansion of

$\displaystyle \Big( x+ \frac{1}{x} \Big)^{2n}$ is $\displaystyle \frac{1\cdot 3\cdot 5 \ldots (2n-1)}{n!} \cdot 2^n$ .

Answer:

$\displaystyle \text{Given expression: } \Big( x+ \frac{1}{x} \Big)^{2n}$

$\displaystyle T_{r+1} = ^{2n} C_r (x)^{2n-r} \Big( \frac{1}{x} \Big)^r$

$\displaystyle = ^{2n} C_r x^{2n-2r}$

If the term is independent of $\displaystyle x , \text{ then }$

$\displaystyle 2n-2r = 0 \Rightarrow r = n$

$\displaystyle \therefore T_{r+1} = ^{2n} C_n = \frac{(2n)!}{ n! n!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots (2n-2) \cdot (2n-1) \cdot (2n)}{n! n!}$

$\displaystyle = \frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)][ 2 \cdot 4 \cdot 6 \cdot 8 \ldots (2n-2) \cdot (2n)}{n! n!}$

$\displaystyle = \frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)] 2^n [ 1 \cdot 2 \cdot 3 \cdot 4 \ldots (n-1) \cdot (n)]}{n! n!}$

$\displaystyle = \frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)] 2^n}{n!}$

Therefore the term independent of $\displaystyle x$ is $\displaystyle \frac{[ 1 \cdot 3 \cdot 5 \cdot 7 \ldots (2n-1)] 2^n}{n!}$

$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{The coefficients of the }5^{\text{th}},6^{\text{th}}\text{ and }7^{\text{th}}\text{ terms in the}$
$\displaystyle \text{expansion of }(1+x)^n\text{ are in A.P. Find }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(1+x)^n$
$\displaystyle \text{Coefficient of the }5^{\text{th}}\text{ term}={^n\mathrm{C}_4}$
$\displaystyle \text{Coefficient of the }6^{\text{th}}\text{ term}={^n\mathrm{C}_5}$
$\displaystyle \text{Coefficient of the }7^{\text{th}}\text{ term}={^n\mathrm{C}_6}$
$\displaystyle \text{Since these coefficients are in A.P.,}$
$\displaystyle 2{^n\mathrm{C}_5}={^n\mathrm{C}_4}+{^n\mathrm{C}_6}$
$\displaystyle \text{Dividing throughout by }{^n\mathrm{C}_5},$
$\displaystyle 2=\frac{{^n\mathrm{C}_4}}{{^n\mathrm{C}_5}}+\frac{{^n\mathrm{C}_6}}{{^n\mathrm{C}_5}}$
$\displaystyle 2=\frac{5}{n-4}+\frac{n-5}{6}$
$\displaystyle 12(n-4)=30+(n-5)(n-4)$
$\displaystyle 12n-48=30+n^2-9n+20$
$\displaystyle 12n-48=n^2-9n+50$
$\displaystyle n^2-21n+98=0$
$\displaystyle (n-7)(n-14)=0$
$\displaystyle \Rightarrow n=7\text{ or }n=14$
$\displaystyle \text{Since }n\geq6,\text{ both values are admissible.}$
$\displaystyle \therefore n=7\text{ or }n=14.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{If the coefficients of the }2^{\text{nd}},3^{\text{rd}}\text{ and }4^{\text{th}}\text{ terms in the}$
$\displaystyle \text{expansion of }(1+x)^{2n}\text{ are in A.P., show that }2n^2-9n+7=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(1+x)^{2n}$
$\displaystyle \text{Coefficient of the }2^{\text{nd}}\text{ term}={^{2n}\mathrm{C}_1}$
$\displaystyle \text{Coefficient of the }3^{\text{rd}}\text{ term}={^{2n}\mathrm{C}_2}$
$\displaystyle \text{Coefficient of the }4^{\text{th}}\text{ term}={^{2n}\mathrm{C}_3}$
$\displaystyle \text{Since these coefficients are in A.P.,}$
$\displaystyle 2{^{2n}\mathrm{C}_2}={^{2n}\mathrm{C}_1}+{^{2n}\mathrm{C}_3}$
$\displaystyle 2\left[\frac{2n(2n-1)}{2}\right]=2n+\frac{2n(2n-1)(2n-2)}{6}$
$\displaystyle 2n(2n-1)=2n+\frac{2n(2n-1)(2n-2)}{6}$
$\displaystyle \text{Dividing throughout by }2n,$
$\displaystyle 2n-1=1+\frac{(2n-1)(2n-2)}{6}$
$\displaystyle 6(2n-1)=6+(2n-1)(2n-2)$
$\displaystyle 12n-6=6+4n^2-6n+2$
$\displaystyle 12n-6=4n^2-6n+8$
$\displaystyle 4n^2-18n+14=0$
$\displaystyle \Rightarrow 2n^2-9n+7=0$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{If the coefficients of the }2^{\text{nd}},3^{\text{rd}}\text{ and }4^{\text{th}}\text{ terms in the}$
$\displaystyle \text{expansion of }(1+x)^n\text{ are in A.P., find the value of }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(1+x)^n$
$\displaystyle \text{Coefficient of the }2^{\text{nd}}\text{ term}={^n\mathrm{C}_1}$
$\displaystyle \text{Coefficient of the }3^{\text{rd}}\text{ term}={^n\mathrm{C}_2}$
$\displaystyle \text{Coefficient of the }4^{\text{th}}\text{ term}={^n\mathrm{C}_3}$
$\displaystyle \text{Since these coefficients are in A.P.,}$
$\displaystyle 2{^n\mathrm{C}_2}={^n\mathrm{C}_1}+{^n\mathrm{C}_3}$
$\displaystyle 2\left[\frac{n(n-1)}{2}\right]=n+\frac{n(n-1)(n-2)}{6}$
$\displaystyle n(n-1)=n+\frac{n(n-1)(n-2)}{6}$
$\displaystyle \text{Dividing throughout by }n,$
$\displaystyle n-1=1+\frac{(n-1)(n-2)}{6}$
$\displaystyle 6(n-1)=6+(n-1)(n-2)$
$\displaystyle 6n-6=6+n^2-3n+2$
$\displaystyle n^2-9n+14=0$
$\displaystyle (n-2)(n-7)=0$
$\displaystyle \Rightarrow n=2\text{ or }n=7$
$\displaystyle \text{Since the }4^{\text{th}}\text{ term must exist, }n\geq3.$
$\displaystyle \therefore n=2\text{ is not admissible.}$
$\displaystyle \therefore n=7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{If, in the expansion of }(1+x)^n,\text{ the coefficients of the }p^{\text{th}}\text{ and}$
$\displaystyle q^{\text{th}}\text{ terms are equal, prove that }p+q=n+2,\text{ where }p\ne q.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(1+x)^n$
$\displaystyle T_p={^n\mathrm{C}_{p-1}}x^{p-1}$
$\displaystyle T_q={^n\mathrm{C}_{q-1}}x^{q-1}$
$\displaystyle \text{Since the coefficients of the }p^{\text{th}}\text{ and }q^{\text{th}}\text{ terms are equal,}$
$\displaystyle {^n\mathrm{C}_{p-1}}={^n\mathrm{C}_{q-1}}$
$\displaystyle \text{Since }p\ne q,\text{ we have }p-1\ne q-1.$
$\displaystyle \therefore (p-1)+(q-1)=n$
$\displaystyle \Rightarrow p+q-2=n$
$\displaystyle \Rightarrow p+q=n+2$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Find }a\text{ if the coefficients of }x^2\text{ and }x^3\text{ in the expansion of}$
$\displaystyle (3+ax)^9\text{ are equal.}$
$\displaystyle \text{Answer:}$
$\displaystyle (3+ax)^9={^9\mathrm{C}_0}3^9+{^9\mathrm{C}_1}3^8(ax)+{^9\mathrm{C}_2}3^7(ax)^2$
$\displaystyle \qquad+{^9\mathrm{C}_3}3^6(ax)^3+\cdots$
$\displaystyle \text{Coefficient of }x^2={^9\mathrm{C}_2}3^7a^2$
$\displaystyle \text{Coefficient of }x^3={^9\mathrm{C}_3}3^6a^3$
$\displaystyle \text{Since the coefficients are equal,}$
$\displaystyle {^9\mathrm{C}_2}3^7a^2={^9\mathrm{C}_3}3^6a^3$
$\displaystyle a^2\left({^9\mathrm{C}_2}3^7-{^9\mathrm{C}_3}3^6a\right)=0$
$\displaystyle \text{Therefore, either }a=0$
$\displaystyle \text{or }{^9\mathrm{C}_2}3^7={^9\mathrm{C}_3}3^6a.$
$\displaystyle a=\frac{{^9\mathrm{C}_2}3^7}{{^9\mathrm{C}_3}3^6}$
$\displaystyle =3\cdot\frac{{^9\mathrm{C}_2}}{{^9\mathrm{C}_3}}$
$\displaystyle =3\cdot\frac{3}{7}$
$\displaystyle =\frac{9}{7}$
$\displaystyle \therefore a=0\text{ or }a=\frac{9}{7}.$
$\displaystyle \text{If }a\ne0,\text{ then }a=\frac{9}{7}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Find the coefficient of }a^4\text{ in the product }(1+2a)^4(2-a)^5\text{ using}$
$\displaystyle \text{the binomial theorem.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(1+2a)^4(2-a)^5$
$\displaystyle (1+2a)^4={^4\mathrm{C}_0}+{^4\mathrm{C}_1}(2a)+{^4\mathrm{C}_2}(2a)^2$
$\displaystyle \qquad+{^4\mathrm{C}_3}(2a)^3+{^4\mathrm{C}_4}(2a)^4$
$\displaystyle =1+8a+24a^2+32a^3+16a^4$
$\displaystyle (2-a)^5={^5\mathrm{C}_0}2^5+{^5\mathrm{C}_1}2^4(-a)+{^5\mathrm{C}_2}2^3(-a)^2$
$\displaystyle \qquad+{^5\mathrm{C}_3}2^2(-a)^3+{^5\mathrm{C}_4}2(-a)^4+{^5\mathrm{C}_5}(-a)^5$
$\displaystyle =32-80a+80a^2-40a^3+10a^4-a^5$
$\displaystyle \therefore \text{Coefficient of }a^4$
$\displaystyle =1(10)+8(-40)+24(80)+32(-80)+16(32)$
$\displaystyle =10-320+1920-2560+512$
$\displaystyle =-438$
$\displaystyle \therefore \text{The coefficient of }a^4\text{ is }-438.$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{In the expansion of }(1+x)^n,\text{ the binomial coefficients of three consecutive}$
$\displaystyle \text{terms are respectively }220,495\text{ and }792.\text{ Find the value of }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three consecutive terms be }T_{r-1},T_r\text{ and }T_{r+1}.$
$\displaystyle \text{Their binomial coefficients are }{^n\mathrm{C}_{r-2}},{^n\mathrm{C}_{r-1}}\text{ and }{^n\mathrm{C}_r},\text{ respectively.}$
$\displaystyle \text{Given, }{^n\mathrm{C}_{r-2}}=220,\quad {^n\mathrm{C}_{r-1}}=495,\quad {^n\mathrm{C}_r}=792$
$\displaystyle \frac{{^n\mathrm{C}_{r-2}}}{{^n\mathrm{C}_{r-1}}}=\frac{220}{495}$
$\displaystyle \Rightarrow \frac{r-1}{n-r+2}=\frac{4}{9}$
$\displaystyle \Rightarrow 9(r-1)=4(n-r+2)$
$\displaystyle \Rightarrow 9r-9=4n-4r+8$
$\displaystyle \Rightarrow 4n+17=13r\qquad\text{... ... ... ... ... i)}$
$\displaystyle \text{Also, }\frac{{^n\mathrm{C}_r}}{{^n\mathrm{C}_{r-1}}}=\frac{792}{495}$
$\displaystyle \Rightarrow \frac{n-r+1}{r}=\frac{8}{5}$
$\displaystyle \Rightarrow 5(n-r+1)=8r$
$\displaystyle \Rightarrow 5n-5r+5=8r$
$\displaystyle \Rightarrow 5n+5=13r\qquad\text{... ... ... ... ... ii)}$
$\displaystyle \text{From i) and ii),}$
$\displaystyle 4n+17=5n+5$
$\displaystyle \Rightarrow n=12$
$\displaystyle \therefore n=12.$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{In the expansion of }(1+x)^n,\text{ the coefficients of three consecutive terms}$
$\displaystyle \text{are respectively }56,70\text{ and }56.\text{ Find }n\text{ and the positions of the terms having}$
$\displaystyle \text{these coefficients.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three consecutive terms be }T_r,T_{r+1}\text{ and }T_{r+2}.$
$\displaystyle \text{Their coefficients are }{^n\mathrm{C}_{r-1}},{^n\mathrm{C}_r}\text{ and }{^n\mathrm{C}_{r+1}},\text{ respectively.}$
$\displaystyle \text{Given, }{^n\mathrm{C}_{r-1}}=56,\quad {^n\mathrm{C}_r}=70,\quad {^n\mathrm{C}_{r+1}}=56$
$\displaystyle \text{Since }{^n\mathrm{C}_{r-1}}={^n\mathrm{C}_{r+1}}\text{ and }r-1\ne r+1,$
$\displaystyle (r-1)+(r+1)=n$
$\displaystyle \Rightarrow 2r=n$
$\displaystyle \Rightarrow r=\frac{n}{2}$
$\displaystyle \text{Also, }\frac{{^n\mathrm{C}_{r-1}}}{{^n\mathrm{C}_r}}=\frac{56}{70}$
$\displaystyle \Rightarrow \frac{r}{n-r+1}=\frac{4}{5}$
$\displaystyle \text{Substituting }r=\frac{n}{2},$
$\displaystyle \frac{\frac{n}{2}}{n-\frac{n}{2}+1}=\frac{4}{5}$
$\displaystyle \Rightarrow \frac{\frac{n}{2}}{\frac{n}{2}+1}=\frac{4}{5}$
$\displaystyle \Rightarrow 5n=4n+8$
$\displaystyle \Rightarrow n=8$
$\displaystyle \therefore r=\frac{n}{2}=4$
$\displaystyle \therefore \text{The required terms are }T_4,T_5\text{ and }T_6.$
$\displaystyle \text{Indeed, }{^8\mathrm{C}_3}=56,\quad {^8\mathrm{C}_4}=70,\quad {^8\mathrm{C}_5}=56.$
$\displaystyle \therefore n=8,\text{ and the coefficients occur in the }4^{\text{th}},5^{\text{th}}\text{ and }6^{\text{th}}\text{ terms.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{If the }3^{\text{rd}},4^{\text{th}},5^{\text{th}}\text{ and }6^{\text{th}}\text{ terms in the expansion of}$
$\displaystyle (x+\alpha)^n\text{ are respectively }a,b,c\text{ and }d,\text{ prove that }\frac{b^2-ac}{c^2-bd}=\frac{5a}{3c}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(x+\alpha)^n$
$\displaystyle T_3={^n\mathrm{C}_2}x^{n-2}\alpha^2=a$
$\displaystyle T_4={^n\mathrm{C}_3}x^{n-3}\alpha^3=b$
$\displaystyle T_5={^n\mathrm{C}_4}x^{n-4}\alpha^4=c$
$\displaystyle T_6={^n\mathrm{C}_5}x^{n-5}\alpha^5=d$
$\displaystyle b^2-ac=ab\left(\frac{b}{a}-\frac{c}{b}\right)$
$\displaystyle c^2-bd=bc\left(\frac{c}{b}-\frac{d}{c}\right)$
$\displaystyle \therefore \frac{b^2-ac}{c^2-bd}=\frac{a}{c}\cdot\frac{\frac{b}{a}-\frac{c}{b}}{\frac{c}{b}-\frac{d}{c}}$
$\displaystyle \frac{b}{a}=\frac{{^n\mathrm{C}_3}x^{n-3}\alpha^3}{{^n\mathrm{C}_2}x^{n-2}\alpha^2}$
$\displaystyle =\frac{{^n\mathrm{C}_3}}{{^n\mathrm{C}_2}}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n-2}{3}\cdot\frac{\alpha}{x}$
$\displaystyle \frac{c}{b}=\frac{{^n\mathrm{C}_4}}{{^n\mathrm{C}_3}}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n-3}{4}\cdot\frac{\alpha}{x}$
$\displaystyle \frac{d}{c}=\frac{{^n\mathrm{C}_5}}{{^n\mathrm{C}_4}}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n-4}{5}\cdot\frac{\alpha}{x}$
$\displaystyle \therefore \frac{b}{a}-\frac{c}{b}=\left(\frac{n-2}{3}-\frac{n-3}{4}\right)\frac{\alpha}{x}$
$\displaystyle =\frac{4n-8-3n+9}{12}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n+1}{12}\cdot\frac{\alpha}{x}$
$\displaystyle \frac{c}{b}-\frac{d}{c}=\left(\frac{n-3}{4}-\frac{n-4}{5}\right)\frac{\alpha}{x}$
$\displaystyle =\frac{5n-15-4n+16}{20}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n+1}{20}\cdot\frac{\alpha}{x}$
$\displaystyle \therefore \frac{\frac{b}{a}-\frac{c}{b}}{\frac{c}{b}-\frac{d}{c}}=\frac{\frac{n+1}{12}\cdot\frac{\alpha}{x}}{\frac{n+1}{20}\cdot\frac{\alpha}{x}}$
$\displaystyle =\frac{20}{12}=\frac{5}{3}$
$\displaystyle \therefore \frac{b^2-ac}{c^2-bd}=\frac{a}{c}\cdot\frac{5}{3}$
$\displaystyle =\frac{5a}{3c}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{If }a,b,c\text{ and }d\text{ are respectively the }6^{\text{th}},7^{\text{th}},8^{\text{th}}\text{ and}$
$\displaystyle 9^{\text{th}}\text{ terms in a binomial expansion, prove that }\frac{b^2-ac}{c^2-bd}=\frac{4a}{3c}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the binomial expression be }(x+\alpha)^n.$
$\displaystyle \text{Given, }T_6=a,\quad T_7=b,\quad T_8=c,\quad T_9=d$
$\displaystyle a={^n\mathrm{C}_5}x^{n-5}\alpha^5$
$\displaystyle b={^n\mathrm{C}_6}x^{n-6}\alpha^6$
$\displaystyle c={^n\mathrm{C}_7}x^{n-7}\alpha^7$
$\displaystyle d={^n\mathrm{C}_8}x^{n-8}\alpha^8$
$\displaystyle b^2-ac=ab\left(\frac{b}{a}-\frac{c}{b}\right)$
$\displaystyle c^2-bd=bc\left(\frac{c}{b}-\frac{d}{c}\right)$
$\displaystyle \therefore \frac{b^2-ac}{c^2-bd}=\frac{a}{c}\cdot\frac{\frac{b}{a}-\frac{c}{b}}{\frac{c}{b}-\frac{d}{c}}$
$\displaystyle \frac{b}{a}=\frac{{^n\mathrm{C}_6}}{{^n\mathrm{C}_5}}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n-5}{6}\cdot\frac{\alpha}{x}$
$\displaystyle \frac{c}{b}=\frac{{^n\mathrm{C}_7}}{{^n\mathrm{C}_6}}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n-6}{7}\cdot\frac{\alpha}{x}$
$\displaystyle \frac{d}{c}=\frac{{^n\mathrm{C}_8}}{{^n\mathrm{C}_7}}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n-7}{8}\cdot\frac{\alpha}{x}$
$\displaystyle \therefore \frac{b}{a}-\frac{c}{b}=\left(\frac{n-5}{6}-\frac{n-6}{7}\right)\frac{\alpha}{x}$
$\displaystyle =\frac{7n-35-6n+36}{42}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n+1}{42}\cdot\frac{\alpha}{x}$
$\displaystyle \frac{c}{b}-\frac{d}{c}=\left(\frac{n-6}{7}-\frac{n-7}{8}\right)\frac{\alpha}{x}$
$\displaystyle =\frac{8n-48-7n+49}{56}\cdot\frac{\alpha}{x}$
$\displaystyle =\frac{n+1}{56}\cdot\frac{\alpha}{x}$
$\displaystyle \therefore \frac{\frac{b}{a}-\frac{c}{b}}{\frac{c}{b}-\frac{d}{c}}=\frac{\frac{n+1}{42}\cdot\frac{\alpha}{x}}{\frac{n+1}{56}\cdot\frac{\alpha}{x}}$
$\displaystyle =\frac{56}{42}$
$\displaystyle =\frac{4}{3}$
$\displaystyle \therefore \frac{b^2-ac}{c^2-bd}=\frac{a}{c}\cdot\frac{4}{3}$
$\displaystyle =\frac{4a}{3c}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{If the }6^{\text{th}},7^{\text{th}}\text{ and }8^{\text{th}}\text{ terms in the expansion of}$
$\displaystyle (x+a)^n\text{ are respectively }112,7\text{ and }\frac{1}{4},\text{ find }x,a\text{ and }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(x+a)^n$
$\displaystyle \text{Given, }T_6=112,\quad T_7=7,\quad T_8=\frac{1}{4}$
$\displaystyle T_6={^n\mathrm{C}_5}x^{n-5}a^5=112$
$\displaystyle T_7={^n\mathrm{C}_6}x^{n-6}a^6=7$
$\displaystyle T_8={^n\mathrm{C}_7}x^{n-7}a^7=\frac{1}{4}$
$\displaystyle \frac{T_7}{T_6}=\frac{{^n\mathrm{C}_6}x^{n-6}a^6}{{^n\mathrm{C}_5}x^{n-5}a^5}=\frac{7}{112}$
$\displaystyle \Rightarrow \frac{{^n\mathrm{C}_6}}{{^n\mathrm{C}_5}}\cdot\frac{a}{x}=\frac{1}{16}$
$\displaystyle \Rightarrow \frac{n-5}{6}\cdot\frac{a}{x}=\frac{1}{16}$
$\displaystyle \Rightarrow \frac{a}{x}=\frac{3}{8n-40}\qquad\text{... ... ... ... ... i)}$
$\displaystyle \text{Also, }\frac{T_8}{T_7}=\frac{{^n\mathrm{C}_7}x^{n-7}a^7}{{^n\mathrm{C}_6}x^{n-6}a^6}=\frac{\frac14}{7}$
$\displaystyle \Rightarrow \frac{{^n\mathrm{C}_7}}{{^n\mathrm{C}_6}}\cdot\frac{a}{x}=\frac{1}{28}$
$\displaystyle \Rightarrow \frac{n-6}{7}\cdot\frac{a}{x}=\frac{1}{28}$
$\displaystyle \Rightarrow \frac{a}{x}=\frac{1}{4n-24}\qquad\text{... ... ... ... ... ii)}$
$\displaystyle \text{From i) and ii),}$
$\displaystyle \frac{3}{8n-40}=\frac{1}{4n-24}$
$\displaystyle \Rightarrow 12n-72=8n-40$
$\displaystyle \Rightarrow 4n=32$
$\displaystyle \Rightarrow n=8$
$\displaystyle \therefore \frac{a}{x}=\frac{1}{8}$
$\displaystyle \Rightarrow a=\frac{x}{8}$
$\displaystyle \text{Substituting in }T_6=112,$
$\displaystyle {^8\mathrm{C}_5}x^3\left(\frac{x}{8}\right)^5=112$
$\displaystyle \Rightarrow \frac{56x^8}{8^5}=112$
$\displaystyle \Rightarrow x^8=\frac{112\cdot8^5}{56}$
$\displaystyle \Rightarrow x^8=2\cdot8^5$
$\displaystyle \Rightarrow x^8=65536=4^8$
$\displaystyle \Rightarrow x=4\text{ or }x=-4$
$\displaystyle \text{Since }a=\frac{x}{8},$
$\displaystyle x=4\Rightarrow a=\frac12$
$\displaystyle x=-4\Rightarrow a=-\frac12$
$\displaystyle \therefore (x,a,n)=\left(4,\frac12,8\right)\text{ or }\left(-4,-\frac12,8\right).$
$\displaystyle \text{If }x\text{ and }a\text{ are positive, then }x=4,\ a=\frac12\text{ and }n=8.$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{If the }2^{\text{nd}},3^{\text{rd}}\text{ and }4^{\text{th}}\text{ terms in the expansion of}$
$\displaystyle (x+a)^n\text{ are }240,720\text{ and }1080\text{ respectively, find }x,a\text{ and }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(x+a)^n$
$\displaystyle \text{Given, }T_2=240,\quad T_3=720,\quad T_4=1080$
$\displaystyle T_2={^n\mathrm{C}_1}x^{n-1}a=240$
$\displaystyle T_3={^n\mathrm{C}_2}x^{n-2}a^2=720$
$\displaystyle T_4={^n\mathrm{C}_3}x^{n-3}a^3=1080$
$\displaystyle \frac{T_3}{T_2}=\frac{{^n\mathrm{C}_2}x^{n-2}a^2}{{^n\mathrm{C}_1}x^{n-1}a}=\frac{720}{240}$
$\displaystyle \Rightarrow \frac{{^n\mathrm{C}_2}}{{^n\mathrm{C}_1}}\cdot\frac{a}{x}=3$
$\displaystyle \Rightarrow \frac{n-1}{2}\cdot\frac{a}{x}=3$
$\displaystyle \Rightarrow \frac{a}{x}=\frac{6}{n-1}\qquad\text{... ... ... ... ... i)}$
$\displaystyle \text{Also, }\frac{T_4}{T_3}=\frac{{^n\mathrm{C}_3}x^{n-3}a^3}{{^n\mathrm{C}_2}x^{n-2}a^2}=\frac{1080}{720}$
$\displaystyle \Rightarrow \frac{{^n\mathrm{C}_3}}{{^n\mathrm{C}_2}}\cdot\frac{a}{x}=\frac{3}{2}$
$\displaystyle \Rightarrow \frac{n-2}{3}\cdot\frac{a}{x}=\frac{3}{2}$
$\displaystyle \Rightarrow \frac{a}{x}=\frac{9}{2n-4}\qquad\text{... ... ... ... ... ii)}$
$\displaystyle \text{From i) and ii),}$
$\displaystyle \frac{6}{n-1}=\frac{9}{2n-4}$
$\displaystyle \Rightarrow 12n-24=9n-9$
$\displaystyle \Rightarrow 3n=15$
$\displaystyle \Rightarrow n=5$
$\displaystyle \text{Substituting }n=5\text{ in i),}$
$\displaystyle \frac{a}{x}=\frac{6}{4}=\frac{3}{2}$
$\displaystyle \Rightarrow a=\frac{3x}{2}\qquad\text{... ... ... ... ... iii)}$
$\displaystyle \text{Using }T_2=240,$
$\displaystyle {^5\mathrm{C}_1}x^4a=240$
$\displaystyle \Rightarrow 5x^4\left(\frac{3x}{2}\right)=240$
$\displaystyle \Rightarrow \frac{15}{2}x^5=240$
$\displaystyle \Rightarrow x^5=32$
$\displaystyle \Rightarrow x=2$
$\displaystyle \text{From iii), }a=\frac{3}{2}\times2=3$
$\displaystyle \therefore x=2,\quad a=3,\quad n=5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{Find }a,b\text{ and }n\text{ in the expansion of }(a+b)^n,\text{ if the first three}$
$\displaystyle \text{terms in the expansion are }729,7290\text{ and }30375,\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }(a+b)^n$
$\displaystyle \text{Given, }T_1=729,\quad T_2=7290,\quad T_3=30375$
$\displaystyle T_1={^n\mathrm{C}_0}a^n=729$
$\displaystyle \Rightarrow a^n=729=3^6\qquad\text{... ... ... ... ... i)}$
$\displaystyle T_2={^n\mathrm{C}_1}a^{n-1}b=7290$
$\displaystyle T_3={^n\mathrm{C}_2}a^{n-2}b^2=30375$
$\displaystyle \frac{T_2}{T_1}=\frac{{^n\mathrm{C}_1}a^{n-1}b}{{^n\mathrm{C}_0}a^n}=\frac{7290}{729}$
$\displaystyle \Rightarrow n\frac{b}{a}=10\qquad\text{... ... ... ... ... ii)}$
$\displaystyle \frac{T_3}{T_2}=\frac{{^n\mathrm{C}_2}a^{n-2}b^2}{{^n\mathrm{C}_1}a^{n-1}b}=\frac{30375}{7290}$
$\displaystyle \Rightarrow \frac{n-1}{2}\cdot\frac{b}{a}=\frac{25}{6}\qquad\text{... ... ... ... ... iii)}$
$\displaystyle \text{Dividing ii) by iii),}$
$\displaystyle \frac{n\frac{b}{a}}{\frac{n-1}{2}\frac{b}{a}}=\frac{10}{\frac{25}{6}}$
$\displaystyle \Rightarrow \frac{2n}{n-1}=\frac{12}{5}$
$\displaystyle \Rightarrow 10n=12n-12$
$\displaystyle \Rightarrow 2n=12$
$\displaystyle \Rightarrow n=6$
$\displaystyle \text{From i), }a^6=3^6$
$\displaystyle \Rightarrow a=3\text{ or }a=-3$
$\displaystyle \text{From ii), }6\frac{b}{a}=10$
$\displaystyle \Rightarrow b=\frac{5a}{3}$
$\displaystyle a=3\Rightarrow b=5$
$\displaystyle a=-3\Rightarrow b=-5$
$\displaystyle \therefore (a,b,n)=(3,5,6)\text{ or }(-3,-5,6).$
$\displaystyle \text{If }a\text{ and }b\text{ are positive, then }a=3,\ b=5\text{ and }n=6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{If the term independent of }x\text{ in the expansion of }\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$
$\displaystyle \text{is }405,\text{ find the value of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$
$\displaystyle \text{Let }T_{r+1}\text{ be the term independent of }x.$
$\displaystyle T_{r+1}={^{10}\mathrm{C}_r}(\sqrt{x})^{10-r}\left(-\frac{k}{x^2}\right)^r$
$\displaystyle =(-1)^r{^{10}\mathrm{C}_r}k^r x^{\frac{10-r}{2}-2r}$
$\displaystyle \text{For }T_{r+1}\text{ to be independent of }x,$
$\displaystyle \frac{10-r}{2}-2r=0$
$\displaystyle \Rightarrow 10-r-4r=0$
$\displaystyle \Rightarrow r=2$
$\displaystyle \therefore \text{The term independent of }x\text{ is the }3^{\text{rd}}\text{ term.}$
$\displaystyle T_3=(-1)^2{^{10}\mathrm{C}_2}k^2$
$\displaystyle =45k^2$
$\displaystyle \text{Given, }T_3=405$
$\displaystyle \Rightarrow 45k^2=405$
$\displaystyle \Rightarrow k^2=9$
$\displaystyle \Rightarrow k=\pm3$
$\displaystyle \therefore k=3\text{ or }k=-3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{Find the }6^{\text{th}}\text{ term in the expansion of }\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^n,$
$\displaystyle \text{if the binomial coefficient of the third term from the end is }45.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(y^{\frac{1}{2}}+x^{\frac{1}{3}}\right)^n$
$\displaystyle \text{The third term from the end has the binomial coefficient }{^n\mathrm{C}_2}.$
$\displaystyle \text{Given, }{^n\mathrm{C}_2}=45$
$\displaystyle \Rightarrow \frac{n(n-1)}{2}=45$
$\displaystyle \Rightarrow n(n-1)=90$
$\displaystyle \Rightarrow n^2-n-90=0$
$\displaystyle \Rightarrow (n-10)(n+9)=0$
$\displaystyle \Rightarrow n=10\text{ or }n=-9$
$\displaystyle \text{Since }n\text{ cannot be negative, }n=10.$
$\displaystyle \text{The general term in the expansion is}$
$\displaystyle T_{r+1}={^{10}\mathrm{C}_r}\left(y^{\frac{1}{2}}\right)^{10-r}\left(x^{\frac{1}{3}}\right)^r$
$\displaystyle \text{For the }6^{\text{th}}\text{ term, }r=5.$
$\displaystyle T_6={^{10}\mathrm{C}_5}\left(y^{\frac{1}{2}}\right)^5\left(x^{\frac{1}{3}}\right)^5$
$\displaystyle ={^{10}\mathrm{C}_5}y^{\frac{5}{2}}x^{\frac{5}{3}}$
$\displaystyle =252x^{\frac{5}{3}}y^{\frac{5}{2}}$
$\displaystyle \therefore \text{The }6^{\text{th}}\text{ term is }252x^{\frac{5}{3}}y^{\frac{5}{2}}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{If }p\text{ is a real number and the middle term in the expansion of}$
$\displaystyle \left(\frac{p}{2}+2\right)^8\text{ is }1120,\text{ find }p.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(\frac{p}{2}+2\right)^8$
$\displaystyle \text{The expansion contains }8+1=9\text{ terms.}$
$\displaystyle \therefore \text{The middle term is the }\left(\frac{8}{2}+1\right)^{\text{th}}=5^{\text{th}}\text{ term.}$
$\displaystyle T_5={^8\mathrm{C}_4}\left(\frac{p}{2}\right)^{8-4}(2)^4$
$\displaystyle ={^8\mathrm{C}_4}\left(\frac{p}{2}\right)^4(2)^4$
$\displaystyle =70p^4$
$\displaystyle \text{Given, }T_5=1120$
$\displaystyle \Rightarrow 70p^4=1120$
$\displaystyle \Rightarrow p^4=16$
$\displaystyle \Rightarrow p=\pm2$
$\displaystyle \therefore p=2\text{ or }p=-2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{Find }n\text{ in the binomial expansion of }\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n,$
$\displaystyle \text{if the ratio of the }7^{\text{th}}\text{ term from the beginning to the }7^{\text{th}}\text{ term from the end}$
$\displaystyle \text{is }\frac{1}{6}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$
$\displaystyle \text{The }7^{\text{th}}\text{ term from the end is the }[(n+1)-7+1]^{\text{th}}=(n-5)^{\text{th}}\text{ term}$
$\displaystyle \text{from the beginning.}$
$\displaystyle T_7={^n\mathrm{C}_6}\left(\sqrt[3]{2}\right)^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^6$
$\displaystyle ={^n\mathrm{C}_6}\frac{2^{\frac{n-6}{3}}}{3^2}$
$\displaystyle T_{n-5}={^n\mathrm{C}_{n-6}}\left(\sqrt[3]{2}\right)^6\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}$
$\displaystyle ={^n\mathrm{C}_{n-6}}\frac{2^2}{3^{\frac{n-6}{3}}}$
$\displaystyle \text{Given, }\frac{T_7}{T_{n-5}}=\frac{1}{6}$
$\displaystyle \Rightarrow \frac{{^n\mathrm{C}_6}\frac{2^{\frac{n-6}{3}}}{3^2}}{{^n\mathrm{C}_{n-6}}\frac{2^2}{3^{\frac{n-6}{3}}}}=\frac{1}{6}$
$\displaystyle \text{Since }{^n\mathrm{C}_6}={^n\mathrm{C}_{n-6}},$
$\displaystyle 2^{\frac{n-6}{3}-2}\cdot3^{\frac{n-6}{3}-2}=\frac{1}{6}$
$\displaystyle \Rightarrow 2^{\frac{n-12}{3}}\cdot3^{\frac{n-12}{3}}=\frac{1}{6}$
$\displaystyle \Rightarrow 6^{\frac{n-12}{3}}=6^{-1}$
$\displaystyle \Rightarrow \frac{n-12}{3}=-1$
$\displaystyle \Rightarrow n-12=-3$
$\displaystyle \Rightarrow n=9$
$\displaystyle \therefore n=9.$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{If the }7^{\text{th}}\text{ term from the beginning and the }7^{\text{th}}\text{ term from the end}$
$\displaystyle \text{in the binomial expansion of }\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n\text{ are equal, find }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given expression: }\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$
$\displaystyle \text{The }7^{\text{th}}\text{ term from the end is the }[(n+1)-7+1]^{\text{th}}=(n-5)^{\text{th}}\text{ term}$
$\displaystyle \text{from the beginning.}$
$\displaystyle T_7={^n\mathrm{C}_6}\left(\sqrt[3]{2}\right)^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^6$
$\displaystyle ={^n\mathrm{C}_6}\frac{2^{\frac{n-6}{3}}}{3^2}$
$\displaystyle T_{n-5}={^n\mathrm{C}_{n-6}}\left(\sqrt[3]{2}\right)^6\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}$
$\displaystyle ={^n\mathrm{C}_{n-6}}\frac{2^2}{3^{\frac{n-6}{3}}}$
$\displaystyle \text{Since }T_7=T_{n-5},$
$\displaystyle {^n\mathrm{C}_6}\frac{2^{\frac{n-6}{3}}}{3^2}={^n\mathrm{C}_{n-6}}\frac{2^2}{3^{\frac{n-6}{3}}}$
$\displaystyle \text{Since }{^n\mathrm{C}_6}={^n\mathrm{C}_{n-6}},$
$\displaystyle \frac{2^{\frac{n-6}{3}}}{3^2}=\frac{2^2}{3^{\frac{n-6}{3}}}$
$\displaystyle \Rightarrow 2^{\frac{n-6}{3}}3^{\frac{n-6}{3}}=2^2\cdot3^2$
$\displaystyle \Rightarrow 6^{\frac{n-6}{3}}=6^2$
$\displaystyle \Rightarrow \frac{n-6}{3}=2$
$\displaystyle \Rightarrow n-6=6$
$\displaystyle \Rightarrow n=12$
$\displaystyle \therefore n=12.$
$\displaystyle \\$