Class 11: Binomial Theorem – Exercise 18.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: Find the 11th term from the beginning and the 11th term from} \\ \\ \text{the end in the expansion of } \Big( 2x- \frac{1}{x^2} \Big)^{25}$ .

Answer:

$\displaystyle \text{Given expression: } \Big( 2x- \frac{1}{x^2} \Big)^{25}$

$\displaystyle \text{Therefore there are } 26 \text{ terms in the expansion. }$

$\displaystyle 11^{th}$ term from the end would be $\displaystyle (26-11+1)$ i.e. $\displaystyle 16^{th}$ term from the beginning for expression $\displaystyle \Big( 2x- \frac{1}{x^2} \Big)^{25}$

$\displaystyle \text{We know that for } (x+a)^n : T_{r+1} = ^{n} C_{r} (x)^{n-r} a^{r}$

$\displaystyle \text{Therefore } T_{16} = T_{15+1} = ^{25} C_{15} (2x)^{25-15} \Big(- \frac{1}{x^2} \Big)^{15}$

$\displaystyle = ^{25} C_{15} (2x)^{10} \Big(- \frac{1}{x^{30}} \Big)$

$\displaystyle = - ^{25} C_{15} \Big( \frac{2^{10}}{x^{20}} \Big)$

$\displaystyle \text{Also } T_{11} = T_{10+1} = ^{25} C_{10} (2x)^{25-10} \Big(- \frac{1}{x^2} \Big)^{10}$

$\displaystyle = ^{25} C_{10} (2x)^{15} \Big( \frac{1}{x^{20}} \Big)$

$\displaystyle = ^{25} C_{10} \Big( \frac{2^{15}}{x^{5}} \Big)$

$\displaystyle \\$

$\displaystyle \text{Question 2: Find the 7th term in the expansion } \Big( 3x^2-\frac{1}{x^3} \Big)^{10}$ .

Answer:

$\displaystyle \text{Given expression: } \Big( 3x^2- \frac{1}{x^3} \Big)^{10}$

$\displaystyle \text{We know that for } (x+a)^n : T_{r+1} = ^{n} C_{r} (x)^{n-r} a^{r}$

$\displaystyle T_{7} = T_{6+1} = ^{10} C_{6} (3x^2)^{10-6} \Big(- \frac{1}{x^3} \Big)^{6}$

$\displaystyle = ^{10} C_{6} \Big( \frac{3^4 x^8}{x^{18}} \Big)$

$\displaystyle = ^{10} C_{6} \Big( \frac{3^4 }{x^{10}} \Big)$

$\displaystyle = \frac{10 \cdot 9 \cdot 8 \cdot 7 }{4 \cdot 3 \cdot 2 \cdot 1 } \times \Big( \frac{3^4}{x^{10}} \Big)$

$\displaystyle = \frac{17010}{x^{10}}$

$\displaystyle \\$

$\displaystyle \text{Question 3: Find the 5th terms from the end in the expansion of } \Big( 3x- \frac{1}{x^2} \Big)^{10}$ .

Answer:

$\displaystyle \text{Given expression: } \Big( 3x- \frac{1}{x^2} \Big)^{10}$

$\displaystyle \text{Therefore there are } 11 \text{ terms in the expansion. }$

$\displaystyle 5^{th}$ term from the end would be $\displaystyle (11-5+1)$ i.e. $\displaystyle 7^{th}$ term from the beginning for expression $\displaystyle \Big( 3x- \frac{1}{x^2} \Big)^{10}$

$\displaystyle T_7 = T_{6+1} = ^{10} C_6 (3x)^{10-6} \Big( \frac{-1}{x^2} \Big)^6$

$\displaystyle = \frac{10 \cdot 9 \cdot 8 \cdot 7 }{4 \cdot 3 \cdot 2 \cdot 1 } 3^4 \cdot x^4 \cdot \frac{1}{x^{12}}$

$\displaystyle = \frac{17010}{x^8}$

$\displaystyle \\$

Question 4: Find the $\displaystyle 8^{th}$ term in the expansion of $\displaystyle ( x^{\frac{3}{2}} y^{\frac{1}{2}} - x^{\frac{1}{2}} y^{\frac{3}{2}} )^{10}$ .

Answer:

$\displaystyle \text{Given expression: } ( x^{\frac{3}{2}} y^{\frac{1}{2}} - x^{\frac{1}{2}} y^{\frac{3}{2}} )^{10}$ .

$\displaystyle T_8 = T_{7+1} = ^{10} C_7 ( x^{\frac{3}{2}} y^{\frac{1}{2}})^{10-7} ( - x^{\frac{1}{2}} y^{\frac{3}{2}} )^7$

$\displaystyle = - \Big( \frac{10 \cdot 9 \cdot 8 }{3 \cdot 2 \cdot 1} \Big) x^{\frac{9}{2}} \cdot y^{\frac{3}{2}} \cdot x^{\frac{7}{2}} \cdot y^{\frac{21}{2}}$

$\displaystyle = - 120 x^8y^{12}$

$\displaystyle \\$

$\displaystyle \text{Question 5: Find the 7th term in the expansion of } \Big( \frac{4x}{5} + \frac{5}{2x} \Big)^8$ .

Answer:

$\displaystyle \text{Given expression: } \Big( \frac{4x}{5} + \frac{5}{2x} \Big)^8$

$\displaystyle T_7 = T_{6+1} = ^8 C_6 \Big( \frac{4x}{5} \Big)^{8-6} \Big( \frac{5}{2x} \Big)^{6}$

$\displaystyle = \frac{8 \cdot 7}{2 \cdot 1} \times \frac{4^2 x^2}{5^2} \times \frac{5^6}{2^6 x^6}$

$\displaystyle = \frac{4375}{x^4}$

$\displaystyle \\$

$\displaystyle \text{Question 6: Find the 4th term in the beginning and the 4th term} \\ \\ \text{ from the end in the expansion } \Big( x + \frac{2}{x} \Big)^9$ .

Answer:

$\displaystyle \text{Given expression: } \Big( x + \frac{2}{x} \Big)^9$

$\displaystyle \text{Therefore there are } 10 \text{ terms in the expansion. }$

$\displaystyle 4^{th}$ term from the end would be $\displaystyle (10-4+1)$ i.e. $\displaystyle 7^{th}$ term from the beginning for expression $\displaystyle \Big( x + \frac{2}{x} \Big)^9$

$\displaystyle T_7 = T_{6+1} =^9 C_6 ( x)^{9-6} \Big( \frac{2}{x} \Big)^6$

$\displaystyle = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} \cdot x^3\cdot \frac{2^6}{x^6}$

$\displaystyle = \frac{5376}{x^3}$

$\displaystyle T_4 = T_{3+1} =^9 C_3 ( x)^{9-3} \Big( \frac{2}{x} \Big)^3$

$\displaystyle = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} \cdot x^6\cdot \frac{2^3}{x^3}$

$\displaystyle = 672x^3$

$\displaystyle \\$

$\displaystyle \text{Question 7: Find the 4th term from the end in the expansion of} \Big( \frac{4x}{5} - \frac{5}{2x} \Big)^9$ .

Answer:

$\displaystyle \text{Given expression: } \Big( \frac{4x}{5} - \frac{5}{2x} \Big)^9$

$\displaystyle \text{Therefore there are } 10 \text{ terms in the expansion. }$

$\displaystyle 4^{th}$ term from the end would be $\displaystyle (10-4+1)$ i.e. $\displaystyle 7^{th}$ term from the beginning for expression $\displaystyle \Big( \frac{4x}{5} - \frac{5}{2x} \Big)^9$

$\displaystyle T_7 = T_{6+1} = ^9 C_6 \Big( \frac{4x}{5} \Big)^{9-6} \Big( \frac{-5}{2x} \Big)^{6}$

$\displaystyle = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} \cdot \frac{4^3x^3}{5^3} \cdot \frac{5^6}{2^6x^6}$

$\displaystyle = \frac{10500}{x^3}$

$\displaystyle \\$

$\displaystyle \text{Question 8: Find the 7th term from the end in the expansion of} \Big( 2x^2 - \frac{3}{2x} \Big)^8$ .

Answer:

$\displaystyle \text{Given expression: } \Big( 2x^2 - \frac{3}{2x} \Big)^8$

$\displaystyle \text{Therefore there are } 10 \text{ terms in the expansion. }$

$\displaystyle 7^{th}$ term from the end would be $\displaystyle (9-7+1)$ i.e. $\displaystyle 3^{rd}$ term from the beginning for expression $\displaystyle \Big( 2x^2 - \frac{3}{2x} \Big)^8$

$\displaystyle T_3 = T_{2+1} = ^8 C_2 (2x^2)^{8-2} \Big( \frac{-3}{2x} \Big)^2$

$\displaystyle = \frac{8 \cdot 7 }{2 \cdot 1} \cdot 2^6 x^{12} \cdot \Big( \frac{3^2}{2^2x^2} \Big)$

$\displaystyle = 4032 x^{10}$

$\displaystyle \\$

Question 9: Find the coefficient of:

$\displaystyle \text{i) } x^{10} \text{ in the expansion of } \Big( 2x^2 - \frac{1}{x} \Big)^{20}$

$\displaystyle \text{ii) } x^{7} \text{ in the expansion of } \Big( x - \frac{1}{x^2} \Big)^{40}$

$\displaystyle \text{iii) } x^{-15} \text{ in the expansion of } \Big( 3x^2 - \frac{a}{3x^3} \Big)^{10}$

$\displaystyle \text{iv) } x^{9} \text{ in the expansion of } \Big( x^2 - \frac{1}{3x} \Big)^{9}$

$\displaystyle \text{v) } x^{m} \text{ in the expansion of } \Big( x + \frac{1}{x} \Big)^{n}$

$\displaystyle \text{vi) } x \text{ in the expansion of } ( 1- 2x^3 + 3x^5) \Big( 1+ \frac{1}{x} \Big)^{8}$

$\displaystyle \text{vii) } a^5b^7 \text{ in the expansion of } (a-2b)^{12}$

$\displaystyle \text{viii) } x \text{ in the expansion of } ( 1 - 3x + 7x^2)( 1 - x)^{16}$

Answer:

$\displaystyle \text{i) } \text{The } \text{Given expression: } \Big( 2x^2 - \frac{1}{x} \Big)^{20}$ .

$\displaystyle \text{Let } (r+1)^{th} \text{ term has } x^{10}$

$\displaystyle T_{r+1} = ^{20} C_{r} \cdot (2x^2)^{20-r} \cdot \Big( \frac{-1}{x} \Big)^r$

$\displaystyle = ^{20} C_{r} \cdot 2^{20-r} \cdot x^{40-2r} \cdot \Big( \frac{-1^r}{x^r} \Big)$

$\displaystyle = (-1)^r \cdot {^{20} C_{r}} \cdot 2^{20-r} \cdot x^{40-3r}$

$\displaystyle \text{Therefore } 40-3r= 10 \Rightarrow r = 10$

$\displaystyle \text{Therefore coefficient of } x^{10} = (-1)^{10} \cdot {^{20} C_{10}} \cdot 2^{20-10} = ^{20} C_{10} \cdot 2^{10}$

$\displaystyle \\$

$\displaystyle \text{ii) } \text{The } \text{Given expression: } \Big( x - \frac{1}{x^2} \Big)^{40}$ .

$\displaystyle \text{Let } (r+1)^{th} \text{ term has } x^{7}$

$\displaystyle T_{r+1} = ^{40} C_{r} \cdot (x)^{40-r} \cdot \Big( \frac{-1}{x^2} \Big)^r$

$\displaystyle = (-1)^r {^{40} C_{r}} \cdot \Big( \frac{x^{40-r}}{x^{2r}} \Big)$

$\displaystyle = (-1)^r \cdot {^{40} C_{r}} \cdot x^{40-3r}$

$\displaystyle \text{Therefore } 40-3r= 7 \Rightarrow r = 11$

$\displaystyle \text{Therefore coefficient of } x^{7} = (-1)^{11} \cdot {^{40} C_{11}} = - ^{40} C_{11}$

$\displaystyle \\$

$\displaystyle \text{iii) } \text{The } \text{Given expression: } \Big( 3x^2 - \frac{a}{3x^3} \Big)^{10}$

$\displaystyle \text{Let } (r+1)^{th} \text{ term has } x^{-15}$

$\displaystyle T_{r+1} = ^{10} C_{r} \cdot (3x^2)^{10-r} \cdot \Big( \frac{-a}{3x^3} \Big)^r$

$\displaystyle = (-1)^r {^{10} C_{r}} \cdot 3^{10-r} x^{20-2r} \Big( \frac{a^r}{3^r x^{3r}} \Big)$

$\displaystyle = (-1)^r \cdot {^{10} C_{r}} 3^{10-2r} a^r \cdot x^{20-5r}$

$\displaystyle \text{Therefore } 20-5r= -15 \Rightarrow r = 7$

$\displaystyle \text{Therefore coefficient of } x^{7} = (-1)^{7} \cdot {^{10} C_{7}} 3^{10-14} a^7 = - ^{10} C_{7} \frac{a^7}{3^4} = \frac{-40}{9} a^7$

$\displaystyle \\$

$\displaystyle \text{iv) } \text{The } \text{Given expression: } \Big( x^2 - \frac{1}{3x} \Big)^{9}$ .

$\displaystyle \text{Let } (r+1)^{th} \text{ term has } x^{9}$

$\displaystyle T_{r+1} = ^{9} C_{r} \cdot (x^2)^{9-r} \cdot \Big( \frac{-1}{3x} \Big)^r$

$\displaystyle = (-1)^r \cdot {^{9} C_{r}} \cdot x^{18-2r} \Big( \frac{1}{3^r x^{r}} \Big)$

$\displaystyle = (-1)^r \cdot {^{9} C_{r}} \cdot \frac{1}{3^r} \cdot x^{18-3r}$

$\displaystyle \text{Therefore } 18-3r= 9 \Rightarrow r = 3$

$\displaystyle \text{Therefore coefficient of } x^{9} = (-1)^{3} \cdot {^{9} C_{3}} \cdot \frac{1}{3^3} = (-1) \cdot \frac{9 \cdot 8 \cdot 7 }{3 \cdot 2 \cdot 1 } \cdot \frac{1}{3^3} = \frac{-28}{9}$

$\displaystyle \\$

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

$\displaystyle \text{Let } (r+1)^{th} \text{ term has } x^{m}$

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

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

$\displaystyle \text{Therefore } n-2r= m \Rightarrow r = \frac{n-m}{2}$

$\displaystyle \text{Therefore coefficient of } x^{m} = ^{n} C_{\frac{n-m}{2}} = \frac{n!}{(\frac{n-m}{2})! (\frac{n+m}{2})!}$

$\displaystyle \\$

$\displaystyle \text{vi) } \text{Given expression: } ( 1- 2x^3 + 3x^5) \Big( 1+ \frac{1}{x} \Big)^{8}$

$\displaystyle = ( 1- 2x^3 + 3x^5) \Big[ ^8 C_0 \Big( \frac{1}{x} \Big)^0 + ^8 C_1 \Big( \frac{1}{x} \Big)^1 +^8 C_2 \Big( \frac{1}{x} \Big)^2 +^8 C_3 \Big( \frac{1}{x} \Big)^3 +^8 C_4 \Big( \frac{1}{x} \Big)^4 \\ \\ { \hspace{5.0cm}+^8 C_5 \Big( \frac{1}{x} \Big)^5 +^8 C_6 \Big( \frac{1}{x} \Big)^6 +^8 C_7 \Big( \frac{1}{x} \Big)^7 + ^8 C_8 \Big( \frac{1}{x} \Big)^8} \Big]$

$\displaystyle = ( 1- 2x^3 + 3x^5) \Big[ ^8 C_0 + ^8 C_1 \Big( \frac{1}{x} \Big) +^8 C_2 \Big( \frac{1}{x^2} \Big) +^8 C_3 \Big( \frac{1}{x^3} \Big) +^8 C_4 \Big( \frac{1}{x^4} \Big) \\ \\ { \hspace{5.0cm}+^8 C_5 \Big( \frac{1}{x^5} \Big) +^8 C_6 \Big( \frac{1}{x^6} \Big) +^8 C_7 \Big( \frac{1}{x^7} \Big) + ^8 C_8 \Big( \frac{1}{x^8} \Big)} \Big]$

$\displaystyle \text{Therefore terms with } x = -2x^3 \cdot ^8 C_2 \cdot \frac{1}{x^2} + 3x^5 \cdot ^8 C_4 \cdot \frac{1}{x^4}$

$\displaystyle = -2 \cdot ^8 C_2 \cdot x + 3 \cdot ^8 C_4 \cdot x$

$\displaystyle = \Big( -2 \cdot \frac{8 \cdot 7}{2 \cdot 1} + 3 \cdot \frac{8 \cdot7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} \Big) x$

$\displaystyle = (-56+210) x$

$\displaystyle = 154x$

Hence the coefficient of the term with $\displaystyle x = 154$

$\displaystyle \\$

$\displaystyle \text{vii) } \text{Given expression: } (a-2b)^{12}$

$\displaystyle = ^{12} C_0 (a)^{12} (-2b)^0 + ^{12} C_1 (a)^{11} (-2b)^1 + ^{12} C_2 (a)^{10} (-2b)^2 \\ \\ { \hspace{3.0cm} + ^{12} C_3 (a)^{9} (-2b)^3 + ^{12} C_4 (a)^{8} (-2b)^4 + ^{12} C_5 (a)^{7} (-2b)^5} \\ \\ {\hspace{3.0cm} + ^{12} C_6 (a)^{6} (-2b)^6 + ^{12} C_7 (a)^{5} (-2b)^7 + \ldots }$

$\displaystyle \text{Therefore coefficient of } a^5b^7 = ^{12} C_7 (-2)^7 = - \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \cdot 2^7 = - 101376$

$\displaystyle \\$

$\displaystyle \text{viii) } \text{Given expression: } ( 1 - 3x + 7x^2)( 1 - x)^{16}$

$\displaystyle = ( 1 - 3x + 7x^2) [ ^{16} C_0 (-x)^0 + ^{16} C_1 (-x)^1 + \ldots ]$

$\displaystyle = ( 1 - 3x + 7x^2) [ 1 + 16(-x) + \ldots ]$

$\displaystyle \text{Therefore coefficient of } x = -3 + ( -16) = - 19$

$\displaystyle \\$

$\displaystyle \text{Question 10: Which term in the expansion of } \Big\{ \Big( \frac{x}{\sqrt{y}} \Big)^{\frac{1}{3}} + \Big( \frac{y}{x^{\frac{1}{3}}} \Big)^{\frac{1}{2}} \Big\}^{21} \text{ contains } \\ \\ x \text{ and } y \text{ to one and the same power? }$

Answer:

$\displaystyle \text{Given expression: } \Big\{ \Big( \frac{x}{\sqrt{y}} \Big)^{\frac{1}{3}} + \Big( \frac{y}{x^{\frac{1}{3}}} \Big)^{\frac{1}{2}} \Big\}^{21}$

$\displaystyle T_{r+1} = ^{21} C_r \cdot \Big[ \Big( \frac{x}{\sqrt{y}} \Big)^{\frac{1}{3}} \Big]^{21-r} \cdot \Big[ \Big( \frac{y}{x^{\frac{1}{3}}} \Big)^{\frac{1}{2}} \Big]^{r}$

$\displaystyle = ^{21} C_r \cdot \Big[ \frac{x^{\frac{21-r}{3}}}{y^{\frac{21-r}{6}}} \Big] \cdot \Big[ \frac{y^{\frac{r}{2}}}{x^{\frac{r}{6}}} \Big]$

$\displaystyle = ^{21} C_r \cdot \frac{ x^{\frac{21-r}{3} - \frac{r}{6}}}{y^{\frac{21-r}{6} - \frac{r}{2}}}$

$\displaystyle = ^{21} C_r \cdot \frac{x^{\frac{42-3r}{6}}}{y^{\frac{21-4r}{6}}}$

$\displaystyle = ^{21} C_r \cdot x^{7 - \frac{r}{2}} \cdot y^{ \frac{2}{3} r - \frac{7}{2}}$

$\displaystyle \text{Therefore } 7 - \frac{r}{2} = \frac{2}{3} r - \frac{7}{2}$

$\displaystyle \frac{21}{2} = \frac{7r}{6} \Rightarrow r = 9$

Therefore the required term $\displaystyle = 10^{th}$ term.

$\displaystyle \\$

$\displaystyle \text{Question 11: Does the expansion of } \Big( 2x^2 - \frac{1}{x} \Big)^{20} \text{ contain any term involving } x^{9}$ .

Answer:

$\displaystyle \text{General expression: } \Big( 2x^2 - \frac{1}{x} \Big)^{20}$

Suppose $\displaystyle x^9$ occurs in the given expression at the $\displaystyle ( r+1)^{th}$ term.

$\displaystyle T_{r+1}= ^{20} C_r \cdot (2x^2)^{20-r} \cdot \Big( \frac{-1}{x} \Big)^r$

$\displaystyle = (-1)^r \cdot {^{20} C_r} \cdot 2^{20-r} \cdot x^{40-2r-r}$

$\displaystyle \text{For the term to contain } x^9$ ,

$\displaystyle 40-3r = 9 \Rightarrow r = \frac{31}{3}$

$\displaystyle \text{Since } r$ can only be an integer, this is not possible. Hence there is no term in the expansion of $\displaystyle \Big( 2x^2 - \frac{1}{x} \Big)^{20}$ that contains $\displaystyle x^9$ .

$\displaystyle \\$

$\displaystyle \text{Question 12: Show that the expansion of } \Big( x^2 + \frac{1}{x} \Big)^{12} \\ \\ \text{ does not contain any term involving } x^{-1}.$

Answer:

$\displaystyle \text{Given expression: } \Big( x^2 + \frac{1}{x} \Big)^{12}$

Suppose $\displaystyle x^{-1}$ occurs in the given expression at the $\displaystyle ( r+1)^{th}$ term.

$\displaystyle T_{r+1}= ^{12} C_r \cdot (x^2)^{12-r} \cdot \Big( \frac{1}{x} \Big)^r$

$\displaystyle = ^{12} C_r \cdot x^{24-2r-r}$

$\displaystyle \text{For the term to contain } x^{-1}$ ,

$\displaystyle 24-3r = -1 \Rightarrow r = \frac{25}{3}$

$\displaystyle \text{Since } r$ can only be an integer, this is not possible. Hence there is no term in the expansion of $\displaystyle \Big( x^2 + \frac{1}{x} \Big)^{12}$ .

$\displaystyle \\$

Question 13: Find the middle term in the expansion of:

$\displaystyle \text{i) } \Big( \frac{2x}{3} - \frac{3}{2x} \Big)^{20} \hspace{1.0cm} \text{ii) } \Big( \frac{a}{x} + bx \Big)^{12} \hspace{1.0cm} \text{iii) } \Big( x^2 - \frac{2}{x} \Big)^{10} \hspace{1.0cm} \text{iv) } \Big( \frac{x}{a} - \frac{a}{x} \Big)^{10}$

Answer:

$\displaystyle \text{i) } \Big( \frac{2x}{3} - \frac{3}{2x} \Big)^{20}$

$\displaystyle \text{Here } n = 20 \text{is even } \text{Therefore } \Big( \frac{20}{2} +1\Big) \text{ i.e. } 11^{th} \text{ term is the middle term. }$

$\displaystyle T_{11} = T_{10+1} = ^{20} C_{10} \cdot \Big( \frac{2}{3} x \Big)^{20-10} \cdot \Big( \frac{-3}{2x} \Big)^{10}$

$\displaystyle = ^{20} C_{10} \cdot \Big( \frac{2}{3} \Big)^{10} x^{10} \cdot \Big( \frac{3}{2} \Big)^{10} \frac{1}{x^{10}}$

$\displaystyle = ^{20} C_{10}$

$\displaystyle \text{ii) } \Big( \frac{a}{x} + bx \Big)^{12}$

$\displaystyle \text{Here } n = 12 \text{is even } \text{Therefore } \Big( \frac{12}{2} +1\Big) \text{ i.e. } 7^{th} \text{ term is the middle term. }$

$\displaystyle T_{7} = T_{6+1} = ^{12} C_{6} \cdot \Big( \frac{a}{x} \Big)^{12-6} \cdot (bx)^6$

$\displaystyle = ^{12} C_{6} a^6b^6$

$\displaystyle = 924 a^6b^6$

$\displaystyle \text{iii) } \Big( x^2 - \frac{2}{x} \Big)^{10}$

$\displaystyle \text{Here } n = 10 \text{is even } \text{Therefore } \Big( \frac{10}{2} +1\Big) \text{ i.e. } 6^{th} \text{ term is the middle term. }$

$\displaystyle T_{6} = T_{5+1} = ^{10} C_{5} \cdot (x^2)^{10-5} \cdot \Big( \frac{-2}{x} \Big)^{5}$

$\displaystyle = - ^{10} C_{5} \cdot x^{10} \cdot \Big( \frac{2^5}{x^5} \Big)$

$\displaystyle = - \frac{ 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \times 2^5 \times x^5$

$\displaystyle = -8064 x^5$

$\displaystyle \text{iv) } \Big( \frac{x}{a} - \frac{a}{x} \Big)^{10}$

$\displaystyle \text{Here } n = 10 \text{is even } \text{Therefore } \Big( \frac{10}{2} +1\Big) \text{ i.e. } 6^{th} \text{ term is the middle term. }$

$\displaystyle T_{6} = T_{5+1} = ^{10} C_{5} \cdot \Big( \frac{x}{a} \Big)^{10-5} \cdot \Big( \frac{-a}{x} \Big)^{5}$

$\displaystyle = - ^{10} C_{5} \cdot \Big( \frac{x^5}{a^5} \Big) \cdot \Big( \frac{a^5}{x^5} \Big)$

$\displaystyle = - ^{10} C_{5} = -252$

$\displaystyle \\$

Question 14: Find the middle term in the expansion of:

$\displaystyle \text{i) } \Big( 3x - \frac{x^3}{6} \Big)^{9} \hspace{1.0cm} \text{ii) } \Big( 2x^2- \frac{1}{x} \Big)^{7} \hspace{1.0cm} \text{iii) } \Big( 3x - \frac{2}{x^2} \Big)^{15} \hspace{1.0cm} \text{iv) } \Big( x^4 - \frac{1}{x^3} \Big)^{11}$

Answer:

$\displaystyle \text{i) } \text{Given expression: } \Big( 3x - \frac{x^3}{6} \Big)^{9}$

$\displaystyle \text{Here } n = 9 \text{is odd } \text{Therefore } \Big( \frac{9+1}{2} \Big) \text{ and } \Big( \frac{9+1}{2} + 1 \Big) \text{ i.e. } 5^{th} \text{ and } 6^{th}$

terms are the middle term of the given expression.

$\displaystyle T_{5} = T_{4+1} = ^{9} C_{4} \cdot (3x)^{9-4} \cdot \Big( \frac{-x^3}{6} \Big)^4$

$\displaystyle = ^{9} C_{4} \cdot 3^5 \cdot x^5 \cdot \frac{x^{12}}{6^4}$

$\displaystyle = \frac{9 \cdot 8 \cdot 7 \cdot 6}{ 4 \cdot 3 \cdot 2 \cdot 1} \times \frac{3^5}{6^4} \times x^{17}$

$\displaystyle = \frac{189}{8} x^{17}$

$\displaystyle T_{6} = T_{5+1} = ^{9} C_{5} \cdot (3x)^{9-5} \cdot \Big( \frac{-x^3}{6} \Big)^5$

$\displaystyle = - ^{9} C_{5} \cdot 3^4 \cdot x^4 \cdot \frac{x^{15}}{6^5}$

$\displaystyle = - \frac{9 \cdot 8 \cdot 7 \cdot 6}{ 4 \cdot 3 \cdot 2 \cdot 1} \times \frac{3^4}{6^5} \times x^{19}$

$\displaystyle = - \frac{21}{16} x^{19}$

$\displaystyle \text{ii) } \text{Given expression: } \Big( 2x^2- \frac{1}{x} \Big)^{7}$

$\displaystyle \text{Here } n = 7 \text{is odd } \text{Therefore } \Big( \frac{7+1}{2} \Big) \text{ and } \Big( \frac{7+1}{2} + 1 \Big) \text{ i.e. } 4^{th} \text{ and } 5^{th}$

terms are the middle term of the given expression.

$\displaystyle T_{4} = T_{3+1} = ^{7} C_{3} \cdot (2x^2)^{7-3} \cdot \Big( \frac{-1}{x} \Big)^3$

$\displaystyle = - ^{7} C_{3} \cdot 2^4 \cdot x^8 \cdot \frac{1}{x^3}$

$\displaystyle = - \frac{7 \cdot 6 \cdot 5}{ 3 \cdot 2 \cdot 1} \times 2^4 \times x^5$

$\displaystyle = - 560x^5$

$\displaystyle T_{5} = T_{4+1} = ^{7} C_{4} \cdot (2x^2)^{7-4} \cdot \Big( \frac{-1}{x} \Big)^4$

$\displaystyle = ^{7} C_{4} \cdot 2^3 \cdot x^6 \cdot \frac{1}{x^4}$

$\displaystyle = \frac{7 \cdot 6 \cdot 5}{ 3 \cdot 2 \cdot 1} \times 2^3 \times x^2$

$\displaystyle = 280x^2$

$\displaystyle \text{iii) } \text{Given expression: } \Big( 3x - \frac{2}{x^2} \Big)^{15}$

$\displaystyle \text{Here } n = 15 \text{is odd } \text{Therefore } \Big( \frac{15+1}{2} \Big) \text{ and } \Big( \frac{15+1}{2} + 1 \Big) \text{ i.e. } 8^{th} \text{ and } 9^{th}$

terms are the middle term of the given expression.

$\displaystyle T_{8} = T_{7+1} = ^{15} C_{7} \cdot (3x)^{15-7} \cdot \Big( \frac{-2}{x^2} \Big)^7$

$\displaystyle = - ^{15} C_{7} \cdot 3^8 \cdot x^8 \cdot \frac{2^7}{x^{14}}$

$\displaystyle = - \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 }{ 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \times 3^8 \times 2^7 \times \frac{1}{x^6}$

$\displaystyle = - \frac{6435 \times 3^8 \times 2^7}{x^6}$

$\displaystyle T_{9} = T_{8+1} = ^{15} C_{8} \cdot (3x)^{15-8} \cdot \Big( \frac{-2}{x^2} \Big)^8$

$\displaystyle = ^{15} C_{8} \cdot 3^7 \cdot x^7 \cdot \frac{2^8}{x^{16}}$

$\displaystyle = \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 }{ 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \times 3^7 \times 2^8 \times \frac{1}{x^9}$

$\displaystyle = \frac{6435 \times 3^7 \times 2^8}{x^9}$

$\displaystyle \text{iv) } \text{Given expression: } \Big( x^4 - \frac{1}{x^3} \Big)^{11}$

$\displaystyle \text{Here } n = 11 \text{is odd } \text{Therefore } \Big( \frac{11+1}{2} \Big) \text{ and } \Big( \frac{11+1}{2} + 1 \Big) \text{ i.e. } 6^{th} \text{ and } 7^{th}$

terms are the middle term of the given expression.

$\displaystyle T_{6} = T_{5+1} = ^{11} C_{5} \cdot (x^4)^{11-5} \cdot \Big( \frac{-1}{x^3} \Big)^5$

$\displaystyle = - ^{11} C_{5} \cdot x^{24} \cdot \frac{1}{x^{15}}$

$\displaystyle = - \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{ 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \times x^9$

$\displaystyle = - 462x^9$

$\displaystyle T_{7} = T_{6+1} = ^{11} C_{6} \cdot (x^4)^{11-6} \cdot \Big( \frac{-1}{x^3} \Big)^6$

$\displaystyle = ^{11} C_{6} \cdot x^{20} \cdot \frac{1}{x^{18}}$

$\displaystyle = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{ 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} x^2$

$\displaystyle = 462x^2$

$\displaystyle \\$

Question 15: Find the middle term in the expansion of:

$\displaystyle \text{i) } \Big( x - \frac{1}{x} \Big)^{10} \hspace{1.0cm} \text{ii) } (1-2x+x^2)^n \hspace{1.0cm} \text{iii) } (1+3x+3x^2+x^3)^{2n}$

$\displaystyle \text{iv) } \Big( 2x - \frac{x^2}{4} \Big)^{9} \hspace{1.0cm} \text{v) } \Big( x - \frac{1}{x} \Big)^{2n+1} \hspace{1.0cm} \text{vi) } \Big( \frac{x}{3} +9y \Big)^{10}$

$\displaystyle \text{vii) } \Big( 3 - \frac{x^3}{6} \Big)^{7} \hspace{1.0cm} \text{viii) } \Big( 2ax - \frac{b}{x^2} \Big)^{12} \hspace{1.0cm} \text{ix) } \Big( \frac{p}{x} + \frac{x}{p} \Big)^{9} \hspace{1.0cm} \text{x) } \Big( \frac{x}{a} - \frac{a}{x} \Big)^{10}$

Answer:

$\displaystyle \text{i) } \text{Given expression: } \Big( x - \frac{1}{x} \Big)^{10}$

$\displaystyle \text{Here } n = 10 \text{ is even } \text{Therefore } \Big( \frac{10}{2} +1\Big) \text{ i.e. } 6^{th} \text{ term is the middle term. }$

$\displaystyle T_6 = T_{5+1} = ^{10} C_5 \cdot (x)^{10-5} \cdot \Big( \frac{-1}{x} \Big)^5$

$\displaystyle = - ^{10} C_5 \cdot \frac{x^5}{x^5}$

$\displaystyle = - \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 }{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

$\displaystyle = - 252$

$\displaystyle \text{ii) } \text{Given expression: } (1-2x+x^2)^n = (1-x)^{2n}$

$\displaystyle \text{Here } 2n \text{ is even } \text{Therefore } \Big( \frac{2n}{2} +1\Big) \text{ i.e. } (n+1)^{th} \text{ term is the middle term. }$

$\displaystyle T_{n+1} = ^{2n} C_n \cdot (1)^{2n-n} \cdot ( -x )^n$

$\displaystyle = (-1)^n \cdot {^{2n} C_n} \cdot x^n$

$\displaystyle = (-1)^n \frac{(2n)!}{(n!)^2} x^n$

$\displaystyle \text{iii) } \text{Given expression: } (1+3x+3x^2+x^3)^{2n} = {(1+x)^3}^{2n} = (1+x)^{6n}$

$\displaystyle \text{Here } 6n \text{ is even } \text{Therefore } \Big( \frac{6n}{2} +1\Big) \text{ i.e. } (3n+1)^{th} \text{ term is the middle term. }$

$\displaystyle T_{3n+1} = ^{6n} C_{3n} \cdot (1)^{6n-3n} \cdot (x)^{3n}$

$\displaystyle = ^{6n} C_{3n} \cdot (x)^{3n}$

$\displaystyle = \frac{(6n)!}{[(3n)!]^2} (x)^{3n}$

$\displaystyle \text{iv) } \text{Given expression: } \Big( 2x - \frac{x^2}{4} \Big)^{9}$

$\displaystyle \text{Here } n = 9 \text{ is odd } \text{Therefore } \Big( \frac{9+1}{2} \Big) \text{ and } \Big( \frac{9+1}{2} + 1 \Big) \text{ i.e. } 5^{th} \text{ and } 6^{th}$

terms are the middle term of the given expression.

$\displaystyle T_5 = T_{4+1} = ^{9} C_4 \cdot (2x)^{9-4} \cdot \Big( \frac{-x^2}{4} \Big)^4$

$\displaystyle = ^{9} C_4 \cdot 2^5 x^5 \cdot \frac{x^8}{4^4}$

$\displaystyle = \frac{9 \cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2\cdot 1} \times \frac{2^5}{4^4} x^{13}$

$\displaystyle = \frac{63}{4} x^{13}$

$\displaystyle T_6 = T_{5+1} = ^{9} C_5 \cdot (2x)^{9-5} \cdot \Big( \frac{-x^2}{4} \Big)^5$

$\displaystyle = - ^{9} C_5 \cdot 2^4 x^4 \cdot \frac{x^{10}}{4^5}$

$\displaystyle = - \frac{9\cdot 8\cdot 7\cdot 6}{4\cdot 3\cdot 2 \cdot 1} \times \frac{2^4}{4^5} x^{14}$

$\displaystyle = - \frac{63}{32} x^{14}$

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

$\displaystyle \text{Here } n = (2n+1) \text{ is odd } \text{Therefore } \Big( \frac{2n+1+1}{2} \Big) \text{ and } \Big( \frac{2n+1+1}{2} + 1 \Big) \text{ i.e. } (n+1)^{th} \text{ and } (n+2)^{th}$

terms are the middle term of the given expression.

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

$\displaystyle = {^{2n+1} C_n} \cdot x^{n+1} \cdot \frac{(-1)^n}{x^n}$

$\displaystyle = (-1)^n \cdot {^{2n+1} C_n} x$

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

$\displaystyle = (-1)^{n+1} \cdot {^{2n+1} C_n} \cdot x^{n} \cdot \frac{1}{x^{n+1}}$

$\displaystyle = (-1)^{n+1} \cdot {^{2n+1} C_n} \cdot \frac{1}{x}$

$\displaystyle \text{vi) } \text{Given expression: } \Big( \frac{x}{3}+9y \Big)^{10}$

$\displaystyle \text{Here } n = 10 \text{ is even } \text{Therefore } \Big( \frac{10}{2} +1\Big) \text{ i.e. } 6^{th} \text{ term is the middle term. }$

$\displaystyle T_6 = T_{5+1} = ^{10} C_5 \cdot \Big( \frac{x}{3} \Big)^{10-5} \cdot (9y)^5$

$\displaystyle = ^{10} C_5 \cdot \frac{x^5}{3^5} \cdot 9^5 \cdot y^5$

$\displaystyle = - \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 }{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{x^5}{3^5} \cdot 9^5 \cdot y^5$

$\displaystyle = 61236 x^5y^5$

$\displaystyle \text{vii) } \text{Given expression: } \Big( 3 - \frac{x^3}{6} \Big)^{7}$

$\displaystyle \text{Here } n = 7 \text{ is odd } \text{Therefore } \Big( \frac{7+1}{2} \Big) \text{ and } \Big( \frac{7+1}{2} + 1 \Big) \text{ i.e. } 4^{th} \text{ and } 5^{th}$

terms are the middle term of the given expression.

$\displaystyle T_4 = T_{3+1} = ^{7} C_7 \cdot (3)^{7-3} \cdot \Big( \frac{-x^3}{6} \Big)^3$

$\displaystyle = (-1) {^{7} C_3} \cdot 3^4 \cdot \frac{x^9}{6^3}$

$\displaystyle = - \frac{105}{8} x^{9}$

$\displaystyle T_5 = T_{4+1} = ^{7} C_4 \cdot (3)^{7-4} \cdot \Big( \frac{-x^3}{6} \Big)^4$

$\displaystyle = {^{7} C_4} \cdot 3^3 \cdot \frac{x^{12}}{6^4}$

$\displaystyle = \frac{35}{48} x^{12}$

$\displaystyle \text{viii) } \text{Given expression: } \Big( 2ax - \frac{b}{x^2} \Big)^{12}$

$\displaystyle \text{Here } n = 12 \text{ is even } \text{Therefore } \Big( \frac{12}{2} +1\Big) \text{ i.e. } 7^{th} \text{ term is the middle term. }$

$\displaystyle T_7 = T_{6+1} = ^{12} C_6 \cdot (2ax)^{12-6} \cdot \Big( \frac{-b}{x^2} \Big)^6$

$\displaystyle = ^{12} C_6 \cdot 2^6 a^6 x^6 \cdot \frac{b^6}{x^{12}}$

$\displaystyle = 59136 \Big( \frac{a^6 b^6 }{x^6} \Big)$

$\displaystyle \text{ix) } \text{Given expression: } \Big( \frac{p}{x} + \frac{x}{p} \Big)^{9}$

$\displaystyle \text{Here } n = 9 \text{ is odd } \text{Therefore } \Big( \frac{9+1}{2} \Big) \text{ and } \Big( \frac{9+1}{2} + 1 \Big) \text{ i.e. } 5^{th} \text{ and } 6^{th}$ terms are the middle term of the given expression.

$\displaystyle T_5 = T_{4+1} = ^{9} C_4 \cdot \Big( \frac{p}{x} \Big)^{9-4} \cdot \Big( \frac{x}{p} \Big)^4$

$\displaystyle = {^{9} C_4} \cdot \Big( \frac{p}{x} \Big)^5 \cdot \Big( \frac{x}{p} \Big)^4$

$\displaystyle = {^{9} C_4} \Big( \frac{p}{x} \Big)$

$\displaystyle = 126 \Big( \frac{p}{x} \Big)$

$\displaystyle T_6 = T_{5+1} = ^{9} C_5 \cdot \Big( \frac{p}{x} \Big)^{9-5} \cdot \Big( \frac{x}{p} \Big)^5$

$\displaystyle = {^{9} C_5} \cdot \Big( \frac{p}{x} \Big)^4 \cdot \Big( \frac{x}{p} \Big)^5$

$\displaystyle = {^{9} C_4} \cdot \Big( \frac{x}{p} \Big)$

$\displaystyle = 126 \Big( \frac{x}{p} \Big)$

$\displaystyle \text{x) } \text{Given expression: } \Big( \frac{x}{a} - \frac{a}{x} \Big)^{10}$

$\displaystyle \text{Here } n = 10 \text{ is even } \text{Therefore } \Big( \frac{10}{2} +1\Big) \text{ i.e. } 6^{th} \text{ term is the middle term. }$

$\displaystyle T_6 = T_{5+1} = ^{10} C_5 \cdot \Big( \frac{x}{a} \Big)^{10-5} \cdot \Big( \frac{-a}{x} \Big)^5$

$\displaystyle = - {^{10} C_5} \cdot \Big( \frac{x}{a} \Big)^5 \cdot \Big( \frac{a}{x} \Big)^5$

$\displaystyle = - {^{10} C_5}$

$\displaystyle = -252$

$\displaystyle \\$

Question 16: Find the term independent of $\displaystyle x$ in the expansion of the following expressions:

$\displaystyle \text{i) } \Big( \frac{3}{2} x^2 - \frac{1}{3x} \Big)^{9} \hspace{1.0cm} \text{ii) } \Big( 2x + \frac{1}{3x^2} \Big)^{9} \hspace{1.0cm} \text{iii) } \Big( 2x^2 - \frac{3}{x^3} \Big)^{25} \\ \\ \\ \text{iv) } \Big( 3x - \frac{2}{x^2} \Big)^{15} \hspace{1.0cm} \text{v) } \Big( \sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{2x^2} \Big)^{10} \hspace{1.0cm} \text{vi) } \Big( x - \frac{1}{x^2} \Big)^{3n} \\ \\ \\ \text{vii) } \Big( \frac{1}{2} x^{\frac{1}{3}} + x^{-\frac{1}{5}} \Big)^{15} \text{viii) } (1 + x + 2x^3) \Big( \frac{3}{2} x^2 - \frac{1}{3x} \Big)^{9} \\ \\ \\ \text{ix) } \Big( \sqrt[3]{x} + \frac{1}{2} \sqrt[3]{x} \Big)^{18} , x > 2 \hspace{1.0cm} \text{x) } \Big( \frac{3}{2} x^2 - \frac{1}{3x} \Big)^{6}$

Answer:

$\displaystyle \text{i) } \text{Given expression: } \Big( \frac{3}{2} x^2 - \frac{1}{3x} \Big)^{9}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle \text{Therefore } T_{r+1} = {^9C_{r}} \cdot \Big( \frac{3}{2} x^2 \Big)^{9-r} \cdot \Big( \frac{-1}{3x} \Big)^{r}$

$\displaystyle = (-1)^r \cdot {^9C_{r}} \cdot \Big( \frac{3}{2} \Big)^{9-r} \cdot \Big( \frac{x^{18-2r}}{3^r x^r} \Big)$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$

$\displaystyle 18-2r-r=0 \Rightarrow r = 6 \text{ Therefore the } 7^{th} \text{ term. }$

$\displaystyle \text{Therefore the required term } \\ \\ T_{6+1} = (-1)^6 \cdot {^9C_{6}} \cdot \Big( \frac{3}{2} \Big)^{9-6} \cdot \frac{1}{3^6} = {^9C_{6}} \cdot \Big( \frac{3^3}{2^3 3^6} \Big) = \frac{7}{18}$

$\displaystyle \\$

$\displaystyle \text{ii) } \text{Given expression: } \Big( 2x + \frac{1}{3x^2} \Big)^{9}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle \text{Therefore } T_{r+1} = {^9C_{r}} \cdot (2x)^{9-r} \cdot \Big( \frac{1}{3x^2} \Big)^{r}$

$\displaystyle = {^9C_{r}} \cdot 2^{9-r} \cdot \Big( \frac{x^{9-r}}{3^r x^{2r}} \Big)$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$

$\displaystyle 9-r-2r=0 \Rightarrow r = 3 \text{ Therefore the } 4^{th} \text{ term. }$

$\displaystyle \text{Therefore the required term } T_{3+1} = ^9 C_3 \cdot \Big( \frac{2^6}{3^3} \Big) = \frac{64}{27} \cdot ^9 C_3 = \frac{1792}{9}$

$\displaystyle \\$

$\displaystyle \text{iii) } \text{Given expression: } \Big( 2x^2 - \frac{3}{x^3} \Big)^{25}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle \text{Therefore } T_{r+1} = {^{25}C_{r}} \cdot (2x^2)^{25-r} \cdot \Big( \frac{-3}{x^3} \Big)^{r}$

$\displaystyle = (-3)^r \cdot {^{25}C_{r}} \cdot 2^{25-r} \cdot \Big( \frac{x^{50-2r}}{x^{3r}} \Big)$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$

$\displaystyle 50-2r-3r=0 \Rightarrow r = 10 \text{ Therefore the } {11}^{th} \text{ term. }$

$\displaystyle \text{Therefore the required term } T_{10+1} =(-3)^{10} \cdot ^{25}C_{10} \cdot 2^{25-10} = 3^{10} \cdot 2^{15} \cdot ^{25}C_{10}$

$\displaystyle \\$

$\displaystyle \text{iv) } \text{Given expression: } \Big( 3x - \frac{2}{x^2} \Big)^{15}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle \text{Therefore } T_{r+1} = {^{15}C_{r}} \cdot (3x)^{15-r} \cdot \Big( \frac{-2}{x^2} \Big)^{r}$

$\displaystyle = (-1)^r \cdot {^{15}C_{r}} \cdot 3^{15-r} \cdot \Big( \frac{x^{15-r}2^r}{x^{2r}} \Big)$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$

$\displaystyle 15-r-2r=0 \Rightarrow r = 5 \text{ Therefore the } {6}^{th} \text{ term. }$

$\displaystyle \text{Therefore the required term } \\ \\ T_{5+1} = ((-1)^5 \cdot {^{15}C_{5}} \cdot 3^{15-5} \cdot 2^5 = - ^{15} C_5 \cdot 3^{10} \cdot 2^5 = - 3003 \cdot 3^{10} \cdot 2^5$

$\displaystyle \\$

$\displaystyle \text{v) } \text{Given expression: } \Big( \sqrt{\frac{x}{3}} + \frac{\sqrt{3}}{2x^2} \Big)^{10}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle \text{Therefore } T_{r+1}= ^{10} C_{r} \cdot ( \sqrt{\frac{x}{3} })^{10-r} \cdot ( \frac{ \sqrt{3} }{2x^2} )^r$

$\displaystyle = {^{10}C_{r}} \cdot \Big( \frac{x^{\frac{10-r}{2} }}{3^{\frac{10-r}{2}} } \cdot \frac{ 3^{ \frac{r}{2} } }{2^r x^{2r}} \Big)$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$

$\displaystyle \frac{10-r}{2} -2r =0 \Rightarrow 10-r-4r = 0 \Rightarrow r = 2 \text{ Therefore the } {3}^{rd} \text{ term. }$

$\displaystyle \text{Therefore the required term } T_{3} = {^{10}C_{2}} \Big( \frac{3^{ \frac{2}{2} }}{2^2 \cdot 3^{\frac{10-2}{2} }} \Big) = {^{10} C_{2}} \Big( \frac{3}{4 \cdot 3^4} \Big) = \frac{5}{12}$

$\displaystyle \\$

$\displaystyle \text{vi) } \text{Given expression: } \Big( x - \frac{1}{x^2} \Big)^{3n}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle T_{r+1} = ^{3n} C_r \cdot (x)^{3n} \cdot \Big( \frac{-1}{x^2} \Big)^r$

$\displaystyle = (-1)^r {^{3n} C_r} \cdot \frac{x^{3n-r}}{x^{2r}}$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$ ,

$\displaystyle 3n-r-2r \Rightarrow r = n$ i.e $\displaystyle (n+1)^{th} \text{ term. }$

$\displaystyle T_{n+1} = ( -1)^n \cdot {^{3n} C_n}$

$\displaystyle \\$

$\displaystyle \text{vii) } \Big( \frac{1}{2} x^{\frac{1}{3}} + x^{-\frac{1}{5}} \Big)^{15}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle T_{r+1} = ^8 C_r \Big( \frac{1}{2} x^{\frac{1}{3}} \Big)^{8-r} \cdot ( x^{\frac{-1}{5}})^r$

$\displaystyle = ^8 C_r \Big( \frac{x^{\frac{8-r}{3}}}{2^{8-r}} \Big) \cdot x^{\frac{-r}{5}}$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$ ,

$\displaystyle \frac{8-r}{3} + \frac{-r}{5} = 0 \Rightarrow 40-5r-3r = 0 \Rightarrow r = 5 \text{Therefore the } {6}^{th} \text{ term. }$

$\displaystyle T_{5+1} = ^8 C_5 \cdot \frac{1}{2^{8-5}} = ^8 C_5 \cdot \frac{1}{8} = 7$

$\displaystyle \\$

$\displaystyle \text{viii) } \text{Given expression: } (1 + x + 2x^3) \Big( \frac{3}{2} x^2 - \frac{1}{3x} \Big)^{9}$

$\displaystyle = (1+x+2x^3) \Big[ ^9 C_0 \Big( \frac{3}{2} x^2 \Big)^{9-0} \Big( \frac{-1}{3x} \Big)^0 + ^9 C_1 \Big( \frac{3}{2} x^2 \Big)^{9-1} \Big( \frac{-1}{3x} \Big)^1 + \ldots + ^9 C_9 \Big( \frac{3}{2}x^2 \Big)^{9-9} \Big( \frac{-1}{3x} \Big)^9\Big]$

$\displaystyle \text{Now } T_{r+1} = ^9 C_r \cdot \Big( \frac{3}{2} x^2 \Big)^{9-r} \cdot \Big( \frac{-1}{3x} \Big)^r$

$\displaystyle = (-1)^r \cdot {^9 C_r} \cdot \Big( \frac{3}{2} \Big)^{9-r} \cdot x^{18-2r} \cdot \Big( \frac{1}{3^r x^r} \Big)$

$\displaystyle = (-1)^r \cdot {^9 C_r} \cdot \Big( \frac{(\frac{3}{2})^{9-r}}{3^r} \Big) \cdot x^{18-2r}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle 18-3r = 0 \Rightarrow r = 6$ . $\displaystyle \text{ Therefore it is the } 7^{th} \text{ term. }$

Coefficient of the $\displaystyle 7^{th}$ term $\displaystyle = (-1)^6 \cdot {^9 C_6} \Big( \frac{(\frac{3}{2})^{9-6}}{3^6} \Big) = \frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1} \times \frac{3^3}{3^6 \cdot 2^3} = \frac{7}{18}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x^{-1}$

$\displaystyle 18-3r = -1 \Rightarrow r = \frac{17}{3}$ . This is not possible. Hence there is no term with $\displaystyle x^{-1}$ in the expansion.

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x^{-3} , \text{ then } ,$

$\displaystyle 18-3r = -3 \Rightarrow r = 7$ . $\displaystyle \text{ Therefore it is the } 8^{th} \text{ term. }$

Coefficient of the $\displaystyle 8^{th}$ term

$\displaystyle = (-1)^7 \cdot {^9 C_7} \Big( \frac{(\frac{3}{2})^{9-7}}{3^7} \Big) = (-1) \frac{9 \cdot 8 }{ 2 \cdot 1} \times \frac{3^2}{3^7 \cdot 2^2} = - \frac{2}{27}$

Therefore the coefficient of the term independent of $\displaystyle x = \frac{7}{18} - \frac{2}{27} = \frac{17}{54}$

$\displaystyle \\$

$\displaystyle \text{ix) } \text{Given expression: } \Big( \sqrt[3]{x} + \frac{1}{2} \sqrt[3]{x} \Big)^{18} , x > 2$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle T_{r+1}= ^{18} C_r ( x^{\frac{1}{3}} )^{18-r} \Big( \frac{1}{2x^{\frac{1}{3}}} \Big)^r$

$\displaystyle = ^{18} C_r \Big( \frac{x^{\frac{18-r}{3}}}{2^r - x^{\frac{r}{3}}} \Big)$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then } ,$

$\displaystyle \frac{18-r}{3} - \frac{r}{3} = 0 \Rightarrow r = 9 \text{ Therefore it is the } {10}^{th}$ term.

$\displaystyle T_{9+1} = ^{18} C_{9} \cdot \frac{1}{2^9} = \frac{ 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{1}{2^9} = \frac{12155}{128}$

$\displaystyle \\$

$\displaystyle \text{x) } \text{Given expression: } \Big( \frac{3}{2} x^2 - \frac{1}{3x} \Big)^{6}$

$\displaystyle \text{Let } T_{r+1}$ term be independent of $\displaystyle x$

$\displaystyle T_{r+1} = ^6C_r \Big( \frac{3}{2} x^2 \Big)^{6-r} \cdot \Big( \frac{-1}{3x} \Big)^r$

$\displaystyle = (-1)^r \cdot {^6C_r} \cdot \Big( \frac{3}{2} \Big)^{6-r} \cdot \Big( \frac{x^{12-2r}}{3^r x^r} \Big)$

$\displaystyle \text{If } T_{r+1} \text{ is to be independent of } x , \text{ then }$ ,

$\displaystyle 12 - 2r - r = 0 \Rightarrow r = 4$ . $\displaystyle \text{ Therefore it is the } {5}^{th}$ term.

$\displaystyle T_{4+1} = (-1)^4 \cdot {^6 C_4} \cdot \Big( \frac{3}{2} \Big)^2 \cdot \frac{1}{3^4} = \frac{6 \cdot 5}{2 \cdot 1} \times \frac{3^2}{2^2} \times \frac{1}{3^4} = \frac{5}{12}$

$\displaystyle \\$

Question 17: If the coefficients of $\displaystyle ( 2r + 4)^{th}$ and $\displaystyle (r-2)^{th}$ terms in the expansion of $\displaystyle ( 1 + x)^{18}$ are equal, find $\displaystyle r$ .

Answer:

$\displaystyle [ \text{Note: When } ^n C_x = ^n C_y \Rightarrow x+y = n ]$

$\displaystyle \text{Given expression: } ( 1 + x)^{18}$

$\displaystyle T_{2r+4} = ^{18} C_{2r+3} \cdot (x)^{2r+3}$

$\displaystyle T_{r-2} = ^{18} C_{r-3} \cdot (x)^{r-3}$

$\displaystyle \text{Given } ^{18} C_{2r+3} = ^{18} C_{r-3}$

$\displaystyle \Rightarrow 2r+3 + r-3 = 18 \Rightarrow 3r = 18 \Rightarrow r = 6$

$\displaystyle \\$

Question 18: If the coefficient of $\displaystyle (2r+1)^{th}$ and $\displaystyle (r + 2)^{th}$ term in the expansion of $\displaystyle ( 1 + x)^{43}$ are equal, find $\displaystyle r$ .

Answer:

$\displaystyle [ \text{Note: When} ^n C_x = ^n C_y \Rightarrow x+y = n ]$

$\displaystyle \text{Given expression: } ( 1 + x)^{43}$

$\displaystyle T_{2r+1} = ^{43} C_{2r} \cdot (x)^{2r}$

$\displaystyle T_{r+2} = ^{43} C_{r+1} \cdot (x)^{r+1}$

$\displaystyle \text{Given } ^{43} C_{2r} = ^{43} C_{r+1}$

$\displaystyle \Rightarrow 2r+r+1 = 43 \Rightarrow 3r = 42 \Rightarrow r = 14$

$\displaystyle \\$

Question 19: Prove that the coefficient of $\displaystyle (r+1)^{th}$ term in the expansion of $\displaystyle ( 1+x)^{n+1}$ is equal to the sum of the coefficients of the $\displaystyle r^{th}$ and $\displaystyle (r+1)^{th}$ terms in the expansion of $\displaystyle ( 1+x)^n$ .

Answer:

For expression: $\displaystyle ( 1+x)^{n+1}$

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

For expression: $\displaystyle (1+x)^n$

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

$\displaystyle T_r = ^n C_{r-1} (x)^{r-1}$

Sum of the coefficients $\displaystyle = ^n C_r + ^n C_{r-1} = ^{n+1} C_r$

Hence proven.

$\displaystyle \\$

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

$\displaystyle \\$

Question 21: The coefficient of the $\displaystyle 5^{th}, 6^{th}$ and $\displaystyle 7^{th}$ terms in the expansion of $\displaystyle (1+x)^n$ are in A.P. , find $\displaystyle n$ .

Answer:

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

Coefficient of $\displaystyle 5^{th} \text{term} = ^n C_{5-1} = ^n C_4$

Coefficient of $\displaystyle 6^{th} \text{term} = ^n C_{6-1} = ^n C_5$

Coefficient of $\displaystyle 7^{th} \text{term} = ^n C_{7-1} = ^n C_6$

Since they are in AP

$\displaystyle 2 \cdot ^n C_5 = ^n C_4 + ^n C_6$

$\displaystyle \Rightarrow 2 \cdot \frac{n!}{5! (n-5)!} = \frac{n!}{4! (n-4)!} + \frac{n!}{6! (n-6)!}$

$\displaystyle \Rightarrow \frac{2}{5 \cdot 4! (n-5) (n-6)!} = \frac{1}{4! (n-4)(n-5) (n-6)!} + \frac{1}{6 \cdot 5 \cdot 4! (n-6)!}$

$\displaystyle \Rightarrow \frac{1}{(n-1)} \Big[ \frac{2}{5} - \frac{1}{(n-4)} \Big] = \frac{1}{30}$

$\displaystyle \Rightarrow \frac{1}{(n-1)} \Big[ \frac{2n - 8 - 5}{5 (n-4)} \Big] = \frac{1}{30}$

$\displaystyle \Rightarrow 6(2n-13) = (n-4)(n-5)$

$\displaystyle \Rightarrow 12n - 78 = n^2 - 9n +20$

$\displaystyle \Rightarrow n^2 - 21 n + 108 = 0$

$\displaystyle \Rightarrow (n-7)(n-14) = 0$

$\displaystyle \Rightarrow n = 7 \ or \ n = 14$

$\displaystyle \\$

Question 22: If the coefficient of $\displaystyle 2^{nd}, 3^{rd}$ and $\displaystyle 4^{th}$ terms in the expansion of $\displaystyle (1+x)^{2n}$ are in A.P., show that $\displaystyle 2n^2 - 9n + 7 = 0$ .

Answer:

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

$\displaystyle T_{1+1} = ^{2n} C_1 \cdot x^1$

$\displaystyle T_{2+1} = ^{2n} C_2 \cdot x^2$

$\displaystyle T_{3+1} = ^{2n} C_3 \cdot x^3$

$\displaystyle \text{Therefore } 2 ( ^{2n} C_2 ) = ^{2n} C_1 + ^{2n} C_3$

$\displaystyle \Rightarrow 2 \cdot \frac{(2n)!}{2! (2n-2)!} = \frac{(2n)!}{1! (2n-1)!} + \frac{(2n)!}{3! (2n-3)!}$

$\displaystyle \Rightarrow \frac{2}{2 (2n-2)(2n-3)!} = \frac{1}{(2n-1)(2n-2)(2n-3)!} + \frac{1}{3 \cdot 2 \cdot 1 ( 2n-3)!}$

$\displaystyle \Rightarrow \frac{1}{2n-2} = \frac{1}{(2n-1)(2n-2)} + \frac{1}{6}$

$\displaystyle \Rightarrow \frac{1}{(2n-2)} \Big[ 1 - \frac{1}{2n-1} \Big] = \frac{1}{6}$

$\displaystyle \Rightarrow \frac{1}{(2n-2)} \Big[ \frac{2n-1-1}{2n-1} \Big] = \frac{1}{6}$

$\displaystyle \Rightarrow 6(2n-2) = ( 2n-2)(2n-1)$

$\displaystyle \Rightarrow 12n - 12 = 4n^2 - 4n - 2n +2$

$\displaystyle \Rightarrow 4n^2 - 18n +14 = 0$

$\displaystyle \Rightarrow 2n^2 -9n+7 = 0$ Hence proved.

$\displaystyle \\$

Question 23: If the coefficient of $\displaystyle 2^{nd}, 3^{rd}$ and $\displaystyle 4^{th}$ terms in the expansion of $\displaystyle (1+x)^{n}$ are in A.P., then find the value on $\displaystyle n$ .

Answer:

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

$\displaystyle T_{1+1} = ^{n} C_1 \cdot x^1$

$\displaystyle T_{2+1} = ^{n} C_2 \cdot x^2$

$\displaystyle T_{3+1} = ^{n} C_3 \cdot x^3$

$\displaystyle \text{Therefore } 2 ( ^{n} C_2 ) = ^{n} C_1 + ^{n} C_3$

$\displaystyle \Rightarrow 2 \cdot \frac{n!}{2! (n-2)!} = \frac{n!}{1! (n-1)!} + \frac{n!}{3! (n-3)!}$

$\displaystyle \Rightarrow \frac{1}{ (n-2)(n-3)!} = \frac{1}{1! (n-1)(n-2)(n-3)!} + \frac{1}{3 \cdot 2 \cdot 1 ( n-3)!}$

$\displaystyle \Rightarrow \frac{1}{n-2} = \frac{1}{(n-1)(n-2)} + \frac{1}{6}$

$\displaystyle \Rightarrow \frac{1}{(n-2)} \Big[ 1 - \frac{1}{n-1} \Big] = \frac{1}{6}$

$\displaystyle \Rightarrow \frac{1}{(n-2)} \Big[ \frac{n-1-1}{n-1} \Big] = \frac{1}{6}$

$\displaystyle \Rightarrow 6(n-2) = ( n-2)(n-1)$

$\displaystyle \Rightarrow (n-2)[ 6 - (n-1)]= 0$

$\displaystyle \Rightarrow (n-2)(7-n) = 0$

$\displaystyle \Rightarrow n = 2 \text{ or } n = 7$

$\displaystyle n = 2$ is not possible as then $\displaystyle 2 > 3$ in the $\displaystyle 4^{th}$ term.

Hence $\displaystyle n = 7$ .

$\displaystyle \\$

Question 24: If in the expansion of $\displaystyle (1+x)^n$ , the coefficients of $\displaystyle p^{th}$ and $\displaystyle q^{th}$ terms are equal, prove the $\displaystyle p+q = n + 2 ,$ where $\displaystyle p \neq q$ .

Answer:

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

$\displaystyle T_p = T _{(p-1)+1} = ^n C_{p-1} x^{p-1}$

$\displaystyle T_q = T _{(q-1)+1} = ^n C_{q-1} x^{q-1}$

$\displaystyle \therefore ^n C_{p-1} = ^n C_{q-1}$

$\displaystyle \Rightarrow (p-1) + (q-1) = n$

$\displaystyle \Rightarrow p+q = n + 2$ . Hence proved.

$\displaystyle \\$

Question 25: Find $\displaystyle a ,$ if the coefficients of $\displaystyle x^2$ and $\displaystyle x^3 \text{ in the expansion of } (3 + ax)^9$ are equal.

Answer:

Given expression:

$\displaystyle (3 + ax)^9= ^9 C_0 (3)^9 (ax)^0 + ^9 C_1 (3)^8 (ax)^1 +^9 C_2 (3)^7 (ax)^2 +^9 C_3 (3)^6 (ax)^3 + \ldots$

$\displaystyle \text{Therefore coefficient of } x^2 = ^9 C_2 (3)^7 a^2$

and coefficient of $\displaystyle x^3 = ^9 C_3 (3)^6 a^3$

$\displaystyle \therefore ^9 C_2 (3)^7 a^2 = ^9 C_3 (3)^6 a^3$

$\displaystyle \frac{9!}{2! 7!} \cdot 3^7 = \frac{9!}{3! 6!} \cdot 3^6 a$

$\displaystyle a = \frac{3\times 3}{7} = \frac{9}{7}$

$\displaystyle \\$

Question 26: Find the coefficient of $\displaystyle a^4$ in the product $\displaystyle (1 + 2a)^4 ( 2 - a)^5$ using binomial theorem.

Answer:

Given expression:

$\displaystyle (1 + 2a)^4 ( 2 - a)^5$

$\displaystyle = \Big[ ^4 C_0 (2a)^0+^4 C_1 (2a)^1+^4 C_2 (2a)^2+^4 C_3 (2a)^3+^4 C_4 (2a)^4 \Big] \times \Big[ ^5 C_0 (2)^5 (-a)^0 \\ \\ + ^5 C_1 (2)^4 (-a)^1 + ^5 C_2 (2)^3 (-a)^2 + ^5 C_3 (2)^2 (-a)^3 + ^5 C_4 (2)^1 (-a)^4 + ^5 C_5 (2)^0 (-a)^5 \Big]$

$\displaystyle = [ 1+8a + 24a^2 + 32a^3 + 16a^4 ] \times [ 32-80a+80a^2 - 40a^3 + 10a^4 - a^5 ]$

$\displaystyle \therefore$ coefficient of $\displaystyle a^4 = 10 + (8)(-40) + (24)(80) + (32)(-80) + (16)(32)$

$\displaystyle = 10 - 320 + 1920 - 2560 + 512$

$\displaystyle = - 438$

$\displaystyle \\$

Question 27: In the expansion of $\displaystyle (1+x)^n$ the binomial coefficients of three consecutive terms are respectively $\displaystyle 220, 495$ and $\displaystyle 792 ,$ find the value of $\displaystyle n$ .

Answer:

Let the consecutive terms be $\displaystyle T_{r-1}, \ \ T_r, \ \ T_{r+1}$

The binomial coefficient for these terms will be $\displaystyle ^n C_{r-2}, \ \ ^n C_{r-1}, \ \ ^n C_{r} ,$ respectively.

It is given, $\displaystyle ^n C_{r-2} = 220, ^n C_{r-1} = 495$ and $\displaystyle ^n C_{r} = 792$

$\displaystyle \therefore \frac{^n C_{r-2}}{^n C_{r-1}} = \frac{220}{495}$

$\displaystyle \Rightarrow \frac{r-1}{n-r+2} = \frac{4}{9}$

$\displaystyle \Rightarrow 9r - 9 = 4n - 4r + 8$

$\displaystyle \Rightarrow 4n + 17 = 13r$ … … … … … i)

$\displaystyle \text{Similarly, } \frac{^n C_{r}}{^n C_{r-1}} = \frac{792}{495}$

$\displaystyle \Rightarrow \frac{n-r+1}{r} = \frac{6}{5}$

$\displaystyle \Rightarrow 5n - 5r + 7 = 8r$

$\displaystyle \Rightarrow 5n + 5 = 13 r$ … … … … … ii)

From i) and ii) we get

$\displaystyle 4n + 17 = 5n + 5$

$\displaystyle \Rightarrow n = 12$

$\displaystyle \\$

Question 28: In the expansion of $\displaystyle (1+x)^n$ the coefficients of three consecutive terms are respectively $\displaystyle 56, 76$ and $\displaystyle 56 ,$ then find $\displaystyle n$ and the position of the terms of these coefficients.

Answer:

Let the consecutive terms be $\displaystyle T_{r}, \ \ T_{r+1}, \ \ T_{r+2}$

The binomial coefficient for these terms will be $\displaystyle ^n C_{r-1}, \ \ ^n C_{r}, \ \ ^n C_{r+1} ,$ respectively.

We have $\displaystyle ^n C_{r-1} = ^n C_{r+1} = 56$

$\displaystyle \Rightarrow r - 1 + r + 1 = n$

$\displaystyle \Rightarrow 2r = n$

$\displaystyle \Rightarrow r = \frac{n}{2}$

Now, $\displaystyle ^n C_{\frac{n}{2}} = 70$ and $\displaystyle ^n C_{\frac{n}{2}-1} = 56$

Dividing $\displaystyle \frac{^n C_{\frac{n}{2}-1}}{^n C_{\frac{n}{2}}} = \frac{56}{70}$

$\displaystyle \Rightarrow \frac{n!}{\big( \frac{n}{2} -1\big) \big( \frac{n}{2} + 1\big) } \times \frac{\big( \frac{n}{2} \big) \big( \frac{n}{2} \big) }{n!} = \frac{8}{10}$

$\displaystyle \Rightarrow \frac{ \big( \frac{n}{2} \big) \big( \frac{n}{2} -1\big) \big( \frac{n}{2} \big) }{ \big(\frac{n}{2}\big) \big( \frac{n}{2} -1\big) \big( \frac{n}{2} + 1 \big)} = \frac{8}{10}$

$\displaystyle \Rightarrow \frac{\frac{n}{2} }{\frac{n}{2}+1} = \frac{8}{10}$

$\displaystyle \Rightarrow 5n = 4n + 8$

$\displaystyle \Rightarrow n = 8$

$\displaystyle \therefore r = \frac{n}{2} = \frac{8}{2} = 4$

Therefore the required terms are $\displaystyle 4^{th}, \ 5^{th}$ and $\displaystyle 6^{th}$ .

$\displaystyle \\$

Question 29: If $\displaystyle 3^{rd}, 4^{th}, 5^{th}$ and $\displaystyle 6^{th}$ terms in the expansion $\displaystyle ( x+ \alpha)^n$ be respectively $\displaystyle a, b, c$ and $\displaystyle d ,$ prove that $\displaystyle \frac{b^2-ac}{c^2-bd} = \frac{5a}{3c}$ .

Answer:

$\displaystyle \text{Given expression: } ( x+ \alpha)^n$

$\displaystyle T_3 = ^n C_2 \ (x)^{n-2} \ (\alpha)^2 T_4 = ^n C_3 \ (x)^{n-3} \ (\alpha)^3$

$\displaystyle T_5 = ^n C_4 \ (x)^{n-4} \ (\alpha)^4 T_6 = ^n C_5 \ (x)^{n-5} \ (\alpha)^5$

Given:

$\displaystyle ^n C_2 \ (x)^{n-2} \ (\alpha)^2 = a ^n C_3 \ (x)^{n-3} (\alpha)^3 = b$

$\displaystyle ^n C_4 \ (x)^{n-4} \ (\alpha)^4 = c ^n C_5 \ (x)^{n-5} (\alpha)^5 = d$

We have to prove:

$\displaystyle \frac{b^2-ac}{c^2-bd} = \frac{5a}{3c}$

$\displaystyle \Rightarrow \frac{b^2 - ac}{a} = \frac{5}{3} \Big[ \frac{c^2- bd}{c} \Big]$

$\displaystyle \Rightarrow \frac{1}{b} \Big[ \frac{b^2 - ac}{a} \Big] = \frac{5}{3} \Big[ \frac{c^2- bd}{c} \Big]$

$\displaystyle \Rightarrow \frac{b}{a} - \frac{c}{b} = \frac{5}{3} \Big[ \frac{c}{d} - \frac{d}{c} \Big]$ … … … … … i)

Substituting the values in i) we get

$\displaystyle \frac{^n C_3 (x)^{n-3} (\alpha)^3}{^n C_2 (x)^{n-2} (\alpha)^2} - \frac{^n C_4 (x)^{n-4} (\alpha)^4}{^n C_3 (x)^{n-3} (\alpha)^3} = \frac{5}{3} \Big[ \frac{^n C_4 (x)^{n-4} (\alpha)^4}{^n C_3 (x)^{n-3} (\alpha)^3} - \frac{^n C_5 (x)^{n-5} (\alpha)^5}{^n C_4 (x)^{n-4} (\alpha)^4} \Big]$

$\displaystyle \Rightarrow \Big[ \frac{^n C_3}{^n C_2} - \frac{^n C_4}{^n C_3} \Big] \frac{ \alpha }{x}= \frac{5 \alpha}{3 x} \Big[ \frac{^n C_4}{^n C_3} - \frac{^n C_5}{^n C_4} \Big]$

We k $\displaystyle \text{Now } \frac{^n C_r}{^n C_{r-1}} = \frac{n-r+1}{r}$

$\displaystyle \Rightarrow \Big[ \frac{n-2}{3} - \frac{n-3}{4} \Big] = \frac{4}{3} \Big[ \frac{n-3}{4} - \frac{n-4}{5} \Big]$

$\displaystyle \Rightarrow \frac{4n-8-3n+9}{3 \times 4} = \frac{5n - 15 - 4n + 16}{3 \times 4}$

$\displaystyle \Rightarrow \frac{n+1}{12} = \frac{n+1}{12}$

Therefore LHS $\displaystyle =$ RHS. Hence proved.

$\displaystyle \\$

Question 30: If $\displaystyle a, b, c$ and $\displaystyle d$ in any binomial expansion be the $\displaystyle 6^{th}, 7^{th}, 8^{th}$ and $\displaystyle 9^{th}$ terms respectively, then prove that $\displaystyle \frac{b^2-ac}{c-bd} = \frac{4a}{3c}$ .

Answer:

Let the expression be $\displaystyle (x + \alpha)^n$

$\displaystyle \text{Given: } T_6 = a, \ \ T_7 = b, \ \ T_8 = c, \ \ T_9 = d$

We have to prove:

$\displaystyle \frac{b^2-ac}{c^2-bd} = \frac{4a}{3c}$

$\displaystyle \Rightarrow \frac{b^2 - ac}{a} = \frac{4}{3} \Big[ \frac{c^2- bd}{c} \Big]$

$\displaystyle \Rightarrow \frac{b}{a} - \frac{c}{b} = \frac{4}{3} \Big[ \frac{c}{d} - \frac{d}{c} \Big]$ … … … … … i)

We know:

$\displaystyle a = ^n C_5 \ x^{n-5} \ {\alpha}^5 b = ^n C_6 \ x^{n-6} \ {\alpha}^6$

$\displaystyle c = ^n C_7 \ x^{n-7} \ {\alpha}^7 d = ^n C_8 \ x^{n-8} \ {\alpha}^8$

Substituting in i) we get

$\displaystyle \frac{^n C_6 \ x^{n-6} \ {\alpha}^6}{^n C_5 \ x^{n-5} \ {\alpha}^5} - \frac{^n C_7 \ x^{n-7} \ {\alpha}^7}{^n C_6 \ x^{n-6} \ {\alpha}^6} = \frac{4}{3} \Big[ \frac{^n C_7 \ x^{n-7} \ {\alpha}^7}{^n C_6 \ x^{n-6} \ {\alpha}^6} - \frac{^n C_8 \ x^{n-8} \ {\alpha}^8}{^n C_7 \ x^{n-7} \ {\alpha}^7} \Big]$

$\displaystyle \Rightarrow \Big[ \frac{^n C_6}{^n C_5} - \frac{^n C_7}{^n C_6} \Big] \frac{ \alpha }{x}= \frac{4 \alpha}{3 x} \Big[ \frac{^n C_7}{^n C_6} - \frac{^n C_8}{^n C_7} \Big]$

We k $\displaystyle \text{Now } \frac{^n C_r}{^n C_{r-1}} = \frac{n-r+1}{r}$

$\displaystyle \Rightarrow \Big[ \frac{n-5}{6} - \frac{n-6}{7} \Big] = \frac{4}{3} \Big[ \frac{n-6}{7} - \frac{n-7}{8} \Big]$

$\displaystyle \Rightarrow \frac{7n-35-6n+36}{6 \times 7} = \frac{8n-48-7n+49}{3 \times 7 \times 2}$

$\displaystyle \Rightarrow \frac{n+1}{42} = \frac{n+1}{42}$

Therefore LHS $\displaystyle =$ RHS. Hence proved.

$\displaystyle \\$

Question 31: If the coefficient of three consecutive terms in the expansion $\displaystyle ( 1+x)^n$ be $\displaystyle 76, 95$ and $\displaystyle 76 ,$ find $\displaystyle n$ .

Answer:

Let the three consecutive terms be $\displaystyle T_r, T_{r+1}, T_{r+2}$

Coefficient of $\displaystyle T_r = ^n C_{r-1} = 76$

Coefficient of $\displaystyle T_{r+1} = ^n C_{r} = 95$

Coefficient of $\displaystyle T_{r+2} = ^n C_{r+1} = 76$

Now, $\displaystyle \frac{^n C_{r+1}}{^n C_{r}} = \frac{76}{95}$

$\displaystyle \Rightarrow \frac{n - ( r+1) + 1}{r+1}= \frac{76}{95}$

$\displaystyle \Rightarrow \frac{n-r}{r+1}= \frac{4}{5}$

$\displaystyle \Rightarrow 5n - 5r = 4r + 4$

$\displaystyle \Rightarrow 5n - 9r = 4$ … … … … … i)

$\displaystyle \text{Similarly, } \frac{^n C_{r}}{^n C_{r-1}} = \frac{95}{76}$

$\displaystyle \Rightarrow \frac{n-r+1}{r}= \frac{5}{4}$

$\displaystyle \Rightarrow 4n - 4r + 4= 5r$

$\displaystyle \Rightarrow 4n-9r=-4$ … … … … … ii)

Subtracting ii) from i) we get $\displaystyle n = 4+4 = 8$

$\displaystyle \\$

Question 32: If the $\displaystyle 6^{th}, 7^{th}$ and $\displaystyle 8^{th} \text{ in the expansion of } (x+a)^n$ are respectively $\displaystyle 112, 7$ and $\displaystyle \frac{1}{4}$ , find $\displaystyle x, a, n$ .

Answer:

$\displaystyle \text{Given expression: } (x+a)^n$

$\displaystyle \text{Given } T_6 = 112, T_7 = 7$ and $\displaystyle T_8 = \frac{1}{4}$

$\displaystyle T_6 = ^n C_5 \ x^{n-5} \ a^5 = 112$

$\displaystyle T_7 = ^n C_6 \ x^{n-6} \ a^6 = 7$

$\displaystyle T_8 = ^n C_7 \ x^{n-7} \ a^7 = \frac{1}{4}$

Now, $\displaystyle \frac{T_7}{T_6} = \frac{^n C_6 \ x^{n-6} \ a^6}{^n C_5 \ x^{n-5} \ a^5} = \frac{7}{112}$

$\displaystyle \Rightarrow \frac{^n C_6}{^n C_5} \times \frac{a}{x} = \frac{1}{16}$

$\displaystyle \Rightarrow \frac{n-5}{6} \times \frac{a}{x} = \frac{1}{16}$

$\displaystyle \Rightarrow \frac{a}{x} = \frac{3}{8n-40}$ … … … … … i)

$\displaystyle \text{Similarly, } \frac{T_8}{T_7} = \frac{^n C_7 \ x^{n-7} \ a^7}{^n C_6 \ x^{n-6} \ a^6} = \frac{\frac{1}{4}}{7}$

$\displaystyle \Rightarrow \frac{^n C_7}{^n C_6} \times \frac{a}{x} = \frac{1}{28}$

$\displaystyle \Rightarrow \frac{n-6}{7} \times \frac{a}{x} = \frac{1}{28}$

$\displaystyle \Rightarrow \frac{a}{x} = \frac{1}{4n-24}$ … … … … … ii)

From i) and ii)

$\displaystyle \frac{3}{8n-40} = \frac{1}{4n-24}$

$\displaystyle \Rightarrow 12n - 72 = 8n -40$

$\displaystyle \Rightarrow 4n = 32$

$\displaystyle \Rightarrow n = 8$

$\displaystyle \text{Therefore } \frac{a}{x} = \frac{3}{24} = \frac{1}{8}$

Substituting we get,

$\displaystyle ^8 C_5 \ x^{8-5} \ \Big( \frac{x}{8} \Big)^5 = 112$

$\displaystyle \Rightarrow \frac{56}{8} x^8 = 112$

$\displaystyle \Rightarrow x^8 = 4^8$

$\displaystyle \Rightarrow x = 4$

Substituting in i) we get $\displaystyle a = \frac{1}{2}$

$\displaystyle \\$

Question 33: If the $\displaystyle 2^{nd}, 3^{rd}$ and $\displaystyle 4^{th}$ term in the expansion of $\displaystyle (x+a)^n$ are $\displaystyle 240, 720$ and $\displaystyle 1080$ respectively, find $\displaystyle x, a, n$ .

Answer:

$\displaystyle \text{Given expression: } (x+a)^n$

$\displaystyle \text{Given } T_2 = 240, T_3 = 720$ and $\displaystyle T_4 = 1080$

$\displaystyle T_2 = ^n C_1 \ x^{n-1} \ a^1 = 240$

$\displaystyle T_3 = ^n C_2 \ x^{n-2} \ a^2 = 720$

$\displaystyle T_4 = ^n C_3 \ x^{n-3} \ a^3 = 1080$

Now, $\displaystyle \frac{T_3}{T_2} = \frac{^n C_2 \ x^{n-2} \ a^2}{^n C_1 \ x^{n-1} \ a^1} = \frac{720}{240}$

$\displaystyle \Rightarrow \frac{n-1}{2x} \cdot a = 3$

$\displaystyle \Rightarrow \frac{a}{x} = \frac{6}{n-1}$ … … … … … i)

$\displaystyle \text{Similarly, } \frac{T_4}{T_3} = \frac{^n C_3 \ x^{n-3} \ a^3}{^n C_2 \ x^{n-2} \ a^2} = \frac{1080}{720}$

$\displaystyle \Rightarrow \frac{n-2}{3x} \cdot a = \frac{3}{2}$

$\displaystyle \Rightarrow \frac{a}{x} = \frac{9}{2n-4}$ … … … … … ii)

From i) and ii) we get

$\displaystyle \frac{6}{n-1} = \frac{9}{2n-4}$

$\displaystyle \Rightarrow 12 n - 24 = 9n - 9$

$\displaystyle \Rightarrow 3n = 15$

$\displaystyle \Rightarrow n = 5$

Substituting in i) we get

$\displaystyle 2a = 3x$ … … … … … iii)

$\displaystyle \text{Now } ^n C_1 \ x^{5-1} \ \Big( \frac{3}{2} x \Big) = 240$

$\displaystyle \Rightarrow 15x^5 = 480$

$\displaystyle \Rightarrow x^5 = 32 = 2^5$

$\displaystyle \Rightarrow x = 2$

From $\displaystyle \text{iii) } a = \frac{3}{2} \times 2 = 3$

$\displaystyle \\$

Question 34: Find $\displaystyle a, b$ and $\displaystyle n$ in the expansion $\displaystyle (a+b)^n ,$ if the first three terms in the expansion are $\displaystyle 729, 7290$ and $\displaystyle 30375 \text{ respectively }$

Answer:

$\displaystyle \text{Given expression: } (a+b)^n$

$\displaystyle \text{Given: } T_1 = 729 \ , T_2 = 7290 \ , T_3 = 30375$

$\displaystyle T_1 = ^n C_0 \ a^n \ b^0 = 729 \Rightarrow a^n = 729 \Rightarrow a^n = 3^6$

$\displaystyle \text{Similarly, } T_2 = ^n C_1 \ a^{n-1} \ b^1 = 7290$

$\displaystyle T_3 = ^n C_2 \ a^{n-2} \ b^2 = 30375$

$\displaystyle \therefore \frac{T_3}{T_2} = \frac{^n C_2 \ a^{n-2} \ b^2}{^n C_1 \ a^{n-1} \ b^1} = \frac{30375}{7290}$

$\displaystyle \Rightarrow \big( \frac{n-1}{2} \big) \big( \frac{b}{a} \big) = \frac{25}{3}$ … … … … … i)

$\displaystyle \text{Similarly, } \therefore \frac{T_2}{T_1} = \frac{^n C_1 \ a^{n-1} \ b^1}{^n C_0 \ a^{n} \ b^0} = \frac{7290}{729}$

$\displaystyle \Rightarrow \big( \frac{nb}{a} \big) = 10$ … … … … … ii)

Dividing ii) by i) we get

$\displaystyle \frac{\frac{nb}{a}}{\frac{(n-1)b}{a}} = 10 \times \frac{3}{25}$

$\displaystyle \Rightarrow \frac{n}{n-1} = \frac{6}{5}$

$\displaystyle \Rightarrow 5n = 6n - 6$

$\displaystyle \Rightarrow n = 6$

$\displaystyle \text{Since } a^6 = 3^6$

$\displaystyle \Rightarrow a = 3$

$\displaystyle \therefore \frac{nb}{a} = 10$

$\displaystyle \Rightarrow b = 10 \times \frac{3}{6} = 5$

$\displaystyle \\$

$\displaystyle \text{Question 35: If the term free from } x \text{ in the expansion } \Big( \sqrt{x} - \frac{k}{x^2} \Big)^{10} \text{ is } \displaystyle 405 , \\ \\ \text{ find the value of } k$ .

Answer:

$\displaystyle \text{Given expression: } \Big( \sqrt{x} - \frac{k}{x^2} \Big)^{10}$

$\displaystyle \text{Let } T_{r+1} \text{ term is independent of } x$

$\displaystyle \text{Therefore } T_{r+1} = ^{10} C_r \cdot (\sqrt{x})^{10-r} \cdot \Big( \frac{-k}{x^2} \Big)^r$

$\displaystyle = (-1)^r \cdot {^{10} C_r} \cdot x^{\frac{10-r}{2}} \cdot \frac{k^r}{x^{2r}}$

$\displaystyle \text{If } T_{r+1} \text{is independent of } x , \text{ then }$

$\displaystyle \frac{10-r}{2} - 2r = 0 \Rightarrow 10-r-4r = 0 \Rightarrow r = 2 . \text{ Therefore it is the } 3^{rd} \text{ term. }$

$\displaystyle \therefore (-1)^2 \cdot {^{10} C_2} \cdot k^2 = 405$

$\displaystyle \Rightarrow \frac{10 \cdot 9}{2 \cdot 1} k^2 = 405$

$\displaystyle \Rightarrow k^2 = \frac{405 \times 2}{10 \times 9} = 9$

$\displaystyle \Rightarrow k = \pm 3$

$\displaystyle \\$

Question 36: Find the $\displaystyle 6^{th}$ term in the expansion $\displaystyle \big( y^{\frac{1}{2}} + x^{\frac{1}{3}} \big)^n ,$ if the binomial coefficient of the third term from the end is $\displaystyle 45$ .

Answer:

$\displaystyle \text{Given expression: } \big( y^{\frac{1}{2}} + x^{\frac{1}{3}}\big)^n$

Third term from the end for $\displaystyle \big( y^{\frac{1}{2}} + x^{\frac{1}{3}} \big)^n$ is third term from the beginning for the expression $\displaystyle \big( x^{\frac{1}{3}} + y^{\frac{1}{2}} \big)^n$

$\displaystyle T_3 = ^n C_2 \cdot \big( x^{\frac{1}{3}} )^{n-2} \cdot ( y^{\frac{1}{2}} \big)^2$

$\displaystyle \text{Given } ^n C_2 = 45$

$\displaystyle \Rightarrow \frac{n(n-1)}{2} = 45$

$\displaystyle \Rightarrow n(n-1) = 90$

$\displaystyle \Rightarrow n^2 - n - 90 = 0$

$\displaystyle \Rightarrow (n-10)(n+9) = 0$

$\displaystyle \Rightarrow n = 10 \text{ or } n = -9$ (not possible as $\displaystyle n$ cannot ne negative)

$\displaystyle \therefore n = 10$

$\displaystyle \text{Therefore } \Rightarrow T_6 = ^{10} C_5 \cdot (y^{\frac{1}{2}})^{10-5} \cdot (x^{\frac{1}{3}})^5$

$\displaystyle \Rightarrow = ^{10} C_5 \cdot y^{\frac{5}{2}} \cdot x^{\frac{5}{3}}$

$\displaystyle \Rightarrow = 252 \cdot y^{\frac{5}{2}} \cdot x^{\frac{5}{3}}$

$\displaystyle \text{Therefore the } \Rightarrow 6^{th}$ term from the beginning is $\displaystyle 252 \cdot y^{\frac{5}{2}} \cdot x^{\frac{5}{3}}$

$\displaystyle \\$

Question 37: If $\displaystyle p$ is a real number and if the middle term in the expansion of $\displaystyle \Big( \frac{p}{2} + 2 \Big)^{8} \text{ is } 1120 , \text{ find } p.$

Answer:

$\displaystyle \text{Given expression: } \Big( \frac{p}{2} + 2 \Big)^{8}$

$\displaystyle \text{Middle term } = \Big( \frac{8}{2} +1 \Big)$ i.e. $\displaystyle 5^{th}$ term

$\displaystyle T_5 = ^8 C_4 \Big( \frac{p}{2} \Big)^{8-4} (2)^4 = 1120$

$\displaystyle \therefore \frac{ 8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} \times \frac{p^4}{2^4} \times 2^4 = 1120$

$\displaystyle \Rightarrow p^4 = \frac{1120}{7 \times 2 \times 5 } = 16 = 2^4$

$\displaystyle \therefore p= \pm 2$

$\displaystyle \\$

$\displaystyle \text{Question 38: Find } n \text{in the binomial } \Big( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \Big)^n , \text{ if the ratio of } 7^{th} \\ \\ \text{ term from the beginning to the } 7^{th} \text{ term from the end is } \frac{1}{6}$ .

Answer:

$\displaystyle \text{Given expression: } \Big( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \Big)^n$

$\displaystyle 7^{th}$ term from the end $\displaystyle = [(n+1)-7+1] = (n-5)^{th}$ term from the beginning.

$\displaystyle T_7 = ^n C_6 \cdot (\sqrt[3]{2})^{n-6} \cdot \Big( \frac{1}{\sqrt[3]{3}} \Big)^6$

$\displaystyle = ^n C_6 \Big( \frac{2^{\frac{n-6}{3}}}{3^2} \Big)$

$\displaystyle T_{n-5} = ^n C_{n-6} \Big(2^{\frac{1}{3}} \Big)^{n - ( n - 6)} \Big( \frac{1}{3^{\frac{1}{3}}} \Big)^{n-6}$

$\displaystyle = ^n C_{n-6} \cdot 2^2 \cdot \Big( \frac{1}{3{\frac{n-6}{3}}} \Big)$

$\displaystyle \text{Given } \frac{T_7}{T_{n-1} } = \frac{1}{6}$

$\displaystyle \Rightarrow \frac{^n C_6 \Big( \frac{2^{\frac{n-6}{3} } }{3^2}\Big) }{^n C_{n-6} \Big( \frac{2^2}{3^{ \frac{n-6}{3} } } \Big) } = \frac{1}{6}$

$\displaystyle \Rightarrow \frac{n!}{6!(n-6)!} \times \frac{(n-6)! 6!}{n!} \cdot 2^{{\frac{n-6}{3}-2} } \cdot 3^{\frac{n-6}{3}-2} =\frac{1}{6}$

$\displaystyle \Rightarrow 2^{\frac{n-12}{3}} \times 3^{ \frac{n-12}{3}} = \frac{1}{6}$

$\displaystyle \Rightarrow 6^{\frac{n-12}{3}} = 6^{-1}$

$\displaystyle \Rightarrow \frac{n-12}{3} = -1$

$\displaystyle \Rightarrow n = 9$

$\displaystyle \\$

Question 39: If the $\displaystyle 7^{th}$ term from the beginning and from the end in the $\displaystyle \text{ binomial expansion } \Big( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \Big)^n \text{ are equal, find } n$ .

Answer:

$\displaystyle \text{Given expression: } \Big( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \Big)^n$

$\displaystyle 7^{th}$ term from the end $\displaystyle = [(n+1)-7+1] = (n-5)^{th}$ term from the beginning.

$\displaystyle T_7 = ^n C_6 \cdot (\sqrt[3]{2})^{n-6} \cdot \Big( \frac{1}{\sqrt[3]{3}} \Big)^6$

$\displaystyle = ^n C_6 \Big( \frac{2^{\frac{n-6}{3}}}{3^2} \Big)$

$\displaystyle T_{n-5} = ^n C_{n-6} \Big(2^{\frac{1}{3}} \Big)^{n - ( n - 6)} \Big( \frac{1}{3^{\frac{1}{3}}} \Big)^{n-6}$

$\displaystyle = ^n C_{n-6} \cdot 2^2 \cdot \Big( \frac{1}{3{\frac{n-6}{3}}} \Big)$

$\displaystyle \text{Since, } T_7 = T_{n-5}$

$\displaystyle \Rightarrow ^n C_6 \Big( \frac{2^{\frac{n-6}{3}}}{3^2} \Big) = ^n C_{n-6} \Big( \frac{2^2}{3^{\frac{n-6}{3}}} \Big)$

$\displaystyle \Rightarrow 6^{\frac{n-6}{3}} = 6^2$

$\displaystyle \Rightarrow \frac{n-6}{3} = 2$

$\displaystyle \Rightarrow n = 12$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 11: Binomial Theorem – Exercise 18.2

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