\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{The number of terms in the expansion of }(x^2-2xy+y^2)^{10}\text{ is:}
\displaystyle \text{(a) }11\qquad\text{(b) }15\qquad\text{(c) }20\qquad\text{(d) }21
\displaystyle \text{Answer:}
\displaystyle x^2-2xy+y^2=(x-y)^2
\displaystyle \therefore (x^2-2xy+y^2)^{10}=\{(x-y)^2\}^{10}=(x-y)^{20}
\displaystyle \text{The expansion of }(x-y)^{20}\text{ contains }20+1=21\text{ terms.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The total number of terms in the expansion of}
\displaystyle (x+a)^{51}-(x-a)^{51}\text{ after simplification is:}
\displaystyle \text{(a) }102\qquad\text{(b) }25\qquad\text{(c) }26\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle (x+a)^{51}=\sum_{r=0}^{51}\binom{51}{r}x^{51-r}a^r
\displaystyle (x-a)^{51}=\sum_{r=0}^{51}\binom{51}{r}x^{51-r}(-a)^r
\displaystyle \text{On subtraction, terms containing even powers of }a\text{ cancel.}
\displaystyle \text{Terms containing odd powers of }a\text{ remain.}
\displaystyle \text{The odd values of }r\text{ from }0\text{ to }51\text{ are }1,3,5,\ldots,51.
\displaystyle \text{Number of such values}=\frac{51+1}{2}=26.
\displaystyle \therefore \text{The simplified expansion contains }26\text{ terms.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The term independent of }x\text{ in the expansion of}
\displaystyle \left(2x+\frac{1}{3x^2}\right)^9\text{ is:}
\displaystyle \text{(a) }2^{\text{nd}}\qquad\text{(b) }3^{\text{rd}}\qquad\text{(c) }4^{\text{th}}\qquad\text{(d) }5^{\text{th}}
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}=\binom{9}{r}(2x)^{9-r}\left(\frac{1}{3x^2}\right)^r
\displaystyle =\binom{9}{r}\frac{2^{9-r}}{3^r}x^{9-3r}
\displaystyle \text{For the term independent of }x,\quad 9-3r=0.
\displaystyle \therefore r=3
\displaystyle \therefore T_{r+1}=T_4.
\displaystyle \therefore \text{The term independent of }x\text{ is the }4^{\text{th}}\text{ term.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The coefficient of }x^{-12}\text{ in the expansion of}
\displaystyle \left(x+\frac{y}{x^3}\right)^{20}\text{ is:}
\displaystyle \text{(a) }\binom{20}{8}\qquad\text{(b) }\binom{20}{8}y^8\qquad\text{(c) }\binom{20}{12}\qquad\text{(d) }\binom{20}{12}y^8
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}=\binom{20}{r}x^{20-r}\left(\frac{y}{x^3}\right)^r
\displaystyle =\binom{20}{r}y^r x^{20-4r}
\displaystyle \text{For the term containing }x^{-12},\quad 20-4r=-12.
\displaystyle 4r=32
\displaystyle r=8
\displaystyle \therefore T_9=\binom{20}{8}y^8x^{-12}
\displaystyle \therefore \text{The coefficient of }x^{-12}\text{ is }\binom{20}{8}y^8.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the binomial expansion of }(a-b)^n,\ n\geq5,\text{ the sum of the }
\displaystyle 5^{\text{th}}\text{ and }6^{\text{th}}\text{ terms is zero. Then }\frac{a}{b}\text{ equals:}
\displaystyle \text{(a) }\frac{n-5}{4}\qquad\text{(b) }\frac{n-4}{5}\qquad\text{(c) }\frac{n-5}{6}\qquad\text{(d) }\frac{n-6}{4}
\displaystyle \text{Answer:}
\displaystyle T_5=\binom{n}{4}a^{n-4}(-b)^4=\binom{n}{4}a^{n-4}b^4
\displaystyle T_6=\binom{n}{5}a^{n-5}(-b)^5=-\binom{n}{5}a^{n-5}b^5
\displaystyle \text{Given, }T_5+T_6=0.
\displaystyle \binom{n}{4}a^{n-4}b^4-\binom{n}{5}a^{n-5}b^5=0
\displaystyle \binom{n}{4}a^{n-4}b^4=\binom{n}{5}a^{n-5}b^5
\displaystyle \therefore \frac{a}{b}=\frac{\binom{n}{5}}{\binom{n}{4}}
\displaystyle =\frac{\frac{n!}{5!(n-5)!}}{\frac{n!}{4!(n-4)!}}
\displaystyle =\frac{n-4}{5}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If the coefficients of the }(2r+3)^{\text{th}}\text{ and }(r-1)^{\text{th}}\text{ terms in}
\displaystyle (1+x)^{15}\text{ are equal, then }r\text{ is:}
\displaystyle \text{(a) }6\qquad\text{(b) }5\qquad\text{(c) }4\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \text{Coefficient of the }(2r+3)^{\text{th}}\text{ term}=\binom{15}{2r+2}
\displaystyle \text{Coefficient of the }(r-1)^{\text{th}}\text{ term}=\binom{15}{r-2}
\displaystyle \text{Given, }\binom{15}{2r+2}=\binom{15}{r-2}.
\displaystyle \text{Since }\binom{n}{p}=\binom{n}{q},\text{ either }p=q\text{ or }p+q=n.
\displaystyle 2r+2=r-2\quad\text{or}\quad(2r+2)+(r-2)=15
\displaystyle r=-4\quad\text{or}\quad3r=15
\displaystyle r=-4\quad\text{or}\quad r=5
\displaystyle \text{Since }r=-4\text{ is not admissible, }r=5.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the expansion of }(1+x)^{12},\text{ the coefficient of the middle term is:}
\displaystyle \text{(a) }308\qquad\text{(b) }462\qquad\text{(c) }924\qquad\text{(d) }1848
\displaystyle \text{Answer:}
\displaystyle \text{The expansion of }(1+x)^{12}\text{ contains }12+1=13\text{ terms.}
\displaystyle \therefore \text{The middle term is the }7^{\text{th}}\text{ term.}
\displaystyle T_7=\binom{12}{6}(1)^{12-6}x^6
\displaystyle =\binom{12}{6}x^6=924x^6
\displaystyle \therefore \text{The coefficient of the middle term is }924.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The coefficient of }x^6\text{ in the expansion of }\left(x^3+\frac{1}{x}\right)^8\text{ is:}
\displaystyle \text{(a) }-252\qquad\text{(b) }252\qquad\text{(c) }63\qquad\text{(d) }-63
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}=\binom{8}{r}(x^3)^{8-r}\left(\frac{1}{x}\right)^r
\displaystyle =\binom{8}{r}x^{24-3r-r}
\displaystyle =\binom{8}{r}x^{24-4r}
\displaystyle \text{For the term containing }x^6,\quad24-4r=6.
\displaystyle 4r=18
\displaystyle r=\frac{9}{2}
\displaystyle \text{Since }r\text{ is not an integer, there is no term containing }x^6.
\displaystyle \therefore \text{The coefficient of }x^6\text{ is }0.
\displaystyle \therefore \text{None of the given options is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If the coefficients of }x^2\text{ and }x^3\text{ in the expansion of }(3+mx)^9
\displaystyle \text{are equal, then the value of }m\text{ is:}
\displaystyle \text{(a) }-\frac{9}{7}\qquad\text{(b) }-\frac{7}{9}\qquad\text{(c) }\frac{9}{7}\qquad\text{(d) }\frac{7}{9}
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}=\binom{9}{r}3^{9-r}(mx)^r.
\displaystyle \text{Coefficient of }x^2=\binom{9}{2}3^7m^2.
\displaystyle \text{Coefficient of }x^3=\binom{9}{3}3^6m^3.
\displaystyle \text{Since the coefficients are equal,}
\displaystyle \binom{9}{2}3^7m^2=\binom{9}{3}3^6m^3.
\displaystyle 36\times3=84m
\displaystyle \therefore m=\frac{108}{84}=\frac{9}{7}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The constant term in }\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}\text{ is }405.
\displaystyle \text{Then }|k|\text{ is:}
\displaystyle \text{(a) }9\qquad\text{(b) }1\qquad\text{(c) }3\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}=\binom{10}{r}(\sqrt{x})^{10-r}\left(-\frac{k}{x^2}\right)^r
\displaystyle =\binom{10}{r}(-k)^r x^{\frac{10-r}{2}-2r}
\displaystyle =\binom{10}{r}(-k)^r x^{5-\frac{5r}{2}}
\displaystyle \text{For the constant term,}\quad5-\frac{5r}{2}=0.
\displaystyle \therefore r=2
\displaystyle \therefore \text{Constant term}=\binom{10}{2}k^2=45k^2.
\displaystyle \text{Given, }45k^2=405.
\displaystyle k^2=9
\displaystyle \therefore |k|=3.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The remainder when }27^{999}\text{ is divided by }7\text{ is:}
\displaystyle \text{(a) }4\qquad\text{(b) }5\qquad\text{(c) }3\qquad\text{(d) }6
\displaystyle \text{Answer:}
\displaystyle 27^{999}=(28-1)^{999}
\displaystyle =\binom{999}{0}28^{999}-\binom{999}{1}28^{998}+\cdots+(-1)^{999}
\displaystyle \text{All the terms except the last term are divisible by }7.
\displaystyle \therefore 27^{999}=7q-1\text{ for some integer }q.
\displaystyle =7(q-1)+6
\displaystyle \therefore \text{The remainder is }6.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The value of }(\sqrt{5}+2)^4+(\sqrt{5}-2)^4\text{ is:}
\displaystyle \text{(a) }212\qquad\text{(b) }322\qquad\text{(c) }218\qquad\text{(d) }328
\displaystyle \text{Answer:}
\displaystyle (\sqrt{5}+2)^4+(\sqrt{5}-2)^4
\displaystyle =2\left[\binom{4}{0}(\sqrt{5})^4+\binom{4}{2}(\sqrt{5})^2(2)^2+\binom{4}{4}(2)^4\right]
\displaystyle =2\left[25+6\times5\times4+16\right]
\displaystyle =2(25+120+16)
\displaystyle =2(161)=322
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }n\in N,\text{ then }2^{2n}-3n-1\text{ is divisible by:}
\displaystyle \text{(a) }3\qquad\text{(b) }9\qquad\text{(c) }4\qquad\text{(d) }16
\displaystyle \text{Answer:}
\displaystyle 2^{2n}=4^n=(1+3)^n
\displaystyle \text{By the Binomial Theorem,}
\displaystyle (1+3)^n=1+3n+\binom{n}{2}3^2+\binom{n}{3}3^3+\cdots
\displaystyle \therefore 2^{2n}-3n-1=\binom{n}{2}3^2+\binom{n}{3}3^3+\cdots
\displaystyle =9\left[\binom{n}{2}+3\binom{n}{3}+3^2\binom{n}{4}+\cdots\right]
\displaystyle \therefore 2^{2n}-3n-1\text{ is divisible by }9.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The }(r+1)^{\text{th}}\text{ term from the end in the expansion of }(x+a)^n\text{ is:}
\displaystyle \text{(a) }\binom{n}{r}x^{n-r}a^r\qquad\text{(b) }\binom{n}{r}x^ra^{n-r}
\displaystyle \text{(c) }\binom{n}{r}x^ra^r\qquad\text{(d) }\binom{n}{r}x^{n-r}a^{n-r}
\displaystyle \text{Answer:}
\displaystyle \text{There are }n+1\text{ terms in the expansion of }(x+a)^n.
\displaystyle \text{The }(r+1)^{\text{th}}\text{ term from the end is the }(n-r+1)^{\text{th}}\text{ term from the beginning.}
\displaystyle T_{n-r+1}=\binom{n}{n-r}x^{n-(n-r)}a^{n-r}
\displaystyle =\binom{n}{r}x^ra^{n-r}
\displaystyle \therefore \text{The }(r+1)^{\text{th}}\text{ term from the end is }\binom{n}{r}x^ra^{n-r}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \text{ASSERTION REASONING QUESTIONS }

\displaystyle \textbf{Question 16: }\text{Assertion }(S_1):\ T_9\text{ is the middle term in the expansion of}
\displaystyle \left(2x+\frac{1}{2x}\right)^8.
\displaystyle \text{Reason }(S_2):\text{ The number of terms in the expansion of }(x+a)^n\text{ is }n+1.
\displaystyle \text{(a) }S_1\text{ is true, }S_2\text{ is true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) }S_1\text{ is true, }S_2\text{ is true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true, }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false, }S_2\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For }\left(2x+\frac{1}{2x}\right)^8,\text{ the number of terms is }8+1=9.
\displaystyle \text{Since there are }9\text{ terms, the middle term is the }5^{\text{th}}\text{ term, i.e., }T_5.
\displaystyle \therefore T_9\text{ is not the middle term. Hence, }S_1\text{ is false.}
\displaystyle \text{Also, the expansion of }(x+a)^n\text{ contains }n+1\text{ terms.}
\displaystyle \therefore S_2\text{ is true.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Assertion }(S_1):\text{ The coefficient of }x^3\text{ in the expansion of }(1+x)^n
\displaystyle \text{is }20,\text{ then }n=6.
\displaystyle \text{Reason }(S_2):\ {}^nC_3=20\Rightarrow n=6.

\displaystyle \text{(a) }S_1\text{ is true, }S_2\text{ is true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) }S_1\text{ is true, }S_2\text{ is true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true, }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false, }S_2\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle \text{In the expansion of }(1+x)^n,\text{ the general term is}
\displaystyle T_{r+1}={}^nC_r x^r.
\displaystyle \therefore \text{the coefficient of }x^3\text{ is }{}^nC_3.
\displaystyle \text{Given, }{}^nC_3=20.
\displaystyle \frac{n(n-1)(n-2)}{6}=20.
\displaystyle n(n-1)(n-2)=120=6\times5\times4.
\displaystyle \therefore n=6.
\displaystyle \therefore S_1\text{ is true.}
\displaystyle \text{Also, }S_2\text{ correctly states that }{}^nC_3=20\Rightarrow n=6.
\displaystyle \therefore S_2\text{ is true and correctly explains }S_1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Assertion }(S_1):\text{ The coefficients of terms equidistant from beginning}
\displaystyle \text{and end in the expansion of }(x+a)^n\text{ are equal.}
\displaystyle \text{Reason }(S_2):\ {}^nC_r={}^nC_{n-r}.

\displaystyle \text{(a) }S_1\text{ is true, }S_2\text{ is true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) }S_1\text{ is true, }S_2\text{ is true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true, }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false, }S_2\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle \text{In the expansion of }(x+a)^n,\text{ the general term is}
\displaystyle T_{r+1}={}^nC_r x^{\,n-r}a^r.
\displaystyle \text{The term equidistant from the other end is}
\displaystyle T_{n-r+1}={}^nC_{n-r}x^r a^{\,n-r}.
\displaystyle \text{We know that }{}^nC_r={}^nC_{n-r}.
\displaystyle \therefore \text{the binomial coefficients of terms equidistant from the beginning and end are equal.}
\displaystyle \therefore S_1\text{ is true and }S_2\text{ is true.}
\displaystyle \text{Also, }S_2\text{ correctly explains }S_1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Assertion }(S_1):\text{ The number of terms in the expansion of}
\displaystyle (3x+y)^8-(3x-y)^8\text{ is }4.
\displaystyle \text{Reason }(S_2):\text{ If }n\text{ is even, then }(x+a)^n-(x-a)^n\text{ has }\frac{n}{2}\text{ terms.}

\displaystyle \text{(a) }S_1\text{ is true, }S_2\text{ is true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) }S_1\text{ is true, }S_2\text{ is true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true, }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false, }S_2\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle \text{In }(3x+y)^8-(3x-y)^8,\text{ the terms containing even powers of }y\text{ cancel.}
\displaystyle \text{Only the terms containing odd powers of }y\text{ remain.}
\displaystyle \text{These correspond to }y^1,y^3,y^5\text{ and }y^7.
\displaystyle \therefore \text{the number of terms is }4.
\displaystyle \therefore S_1\text{ is true.}
\displaystyle \text{In general, for even }n,\ (x+a)^n-(x-a)^n\text{ contains only odd powers of }a.
\displaystyle \text{There are }\frac{n}{2}\text{ such terms.}
\displaystyle \therefore S_2\text{ is true and correctly explains }S_1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Assertion }(S_1):\text{ The }r^{th}\text{ term from the end in the expansion of}
\displaystyle (x+a)^n\text{ is }{}^nC_{n-r+1}x^{r-1}a^{n-r+1}.
\displaystyle \text{Reason }(S_2):\text{ The }r^{th}\text{ term from the end in the expansion of }(x+a)^n
\displaystyle \text{is }(n-r+2)^{th}\text{ term from the beginning.}

\displaystyle \text{(a) }S_1\text{ is true, }S_2\text{ is true and }S_2\text{ is the correct explanation of }S_1.
\displaystyle \text{(b) }S_1\text{ is true, }S_2\text{ is true but }S_2\text{ is not the correct explanation of }S_1.
\displaystyle \text{(c) }S_1\text{ is true, }S_2\text{ is false.}
\displaystyle \text{(d) }S_1\text{ is false, }S_2\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion of }(x+a)^n\text{ is}
\displaystyle T_{k}={}^nC_{k-1}x^{n-k+1}a^{k-1}.
\displaystyle \text{There are }n+1\text{ terms in the expansion.}
\displaystyle \therefore \text{the }r^{th}\text{ term from the end is the }(n-r+2)^{th}\text{ term from the beginning.}
\displaystyle T_{n-r+2}={}^nC_{n-r+1}x^{r-1}a^{n-r+1}.
\displaystyle \therefore S_1\text{ is true and }S_2\text{ is also true.}
\displaystyle \text{Also, }S_2\text{ correctly explains }S_1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 21: }\text{If the coefficients of }2^{nd},\ 3^{rd}\text{ and the }4^{th}\text{ terms in the expansion of}
\displaystyle (1+x)^n\text{ are in A.P. then find the value of }n.
\displaystyle \text{Answer:}
\displaystyle \text{The coefficients of the }2^{nd},\ 3^{rd}\text{ and }4^{th}\text{ terms are}
\displaystyle {}^nC_1,\quad{}^nC_2,\quad{}^nC_3.
\displaystyle \text{Since these are in A.P.,}
\displaystyle 2\,{}^nC_2={}^nC_1+{}^nC_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{Since }n\geq3,\text{ dividing by }n,\text{ we get}
\displaystyle n-1=1+\frac{(n-1)(n-2)}{6}.
\displaystyle n-2=\frac{(n-1)(n-2)}{6}.
\displaystyle 6=n-1.
\displaystyle \therefore n=7.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Prove that: }\sum_{r=0}^{n}{}^nC_r4^r=5^n.
\displaystyle \text{Answer:}
\displaystyle \text{By the Binomial Theorem,}
\displaystyle (x+y)^n=\sum_{r=0}^{n}{}^nC_r x^{n-r}y^r.
\displaystyle \text{Putting }x=1\text{ and }y=4,\text{ we get}
\displaystyle (1+4)^n=\sum_{r=0}^{n}{}^nC_r(1)^{n-r}4^r.
\displaystyle 5^n=\sum_{r=0}^{n}{}^nC_r4^r.
\displaystyle \therefore \sum_{r=0}^{n}{}^nC_r4^r=5^n.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Find }a,\text{ if the }17^{th}\text{ and }18^{th}\text{ terms of the expansion }(2+a)^{50}\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion of }(2+a)^{50}\text{ is}
\displaystyle T_{r+1}={}^{50}C_r2^{50-r}a^r.
\displaystyle \therefore T_{17}={}^{50}C_{16}2^{34}a^{16}.
\displaystyle T_{18}={}^{50}C_{17}2^{33}a^{17}.
\displaystyle \text{Given, }T_{17}=T_{18}.
\displaystyle {}^{50}C_{16}2^{34}a^{16}={}^{50}C_{17}2^{33}a^{17}.
\displaystyle \frac{{}^{50}C_{17}}{{}^{50}C_{16}}\times\frac{a}{2}=1.
\displaystyle \frac{50-16}{17}\times\frac{a}{2}=1.
\displaystyle \frac{34}{17}\times\frac{a}{2}=1.
\displaystyle 2\times\frac{a}{2}=1.
\displaystyle \therefore a=1.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The coefficients of three consecutive terms in the expansion of}
\displaystyle (1+x)^n\text{ are in the ratio }1:7:42.\text{ Find }n.
\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive coefficients be }{}^nC_r,\ {}^nC_{r+1},\ {}^nC_{r+2}.
\displaystyle {}^nC_r:{}^nC_{r+1}:{}^nC_{r+2}=1:7:42.
\displaystyle \therefore \frac{{}^nC_{r+1}}{{}^nC_r}=7.
\displaystyle \frac{n-r}{r+1}=7.
\displaystyle \therefore n-r=7r+7.
\displaystyle \therefore n=8r+7.\qquad ...(1)
\displaystyle \text{Also, }\frac{{}^nC_{r+2}}{{}^nC_{r+1}}=\frac{42}{7}=6.
\displaystyle \frac{n-r-1}{r+2}=6.
\displaystyle \therefore n-r-1=6r+12.
\displaystyle \therefore n=7r+13.\qquad ...(2)
\displaystyle \text{From (1) and (2),}
\displaystyle 8r+7=7r+13.
\displaystyle \therefore r=6.
\displaystyle \therefore n=8(6)+7=55.
\displaystyle \therefore n=55.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Expand: }(x^2+1-2x)^3.
\displaystyle \text{Answer:}
\displaystyle x^2+1-2x=x^2-2x+1=(x-1)^2.
\displaystyle \therefore (x^2+1-2x)^3=\left((x-1)^2\right)^3=(x-1)^6.
\displaystyle \text{By the Binomial Theorem,}
\displaystyle (x-1)^6={}^{6}C_0x^6-{}^{6}C_1x^5+{}^{6}C_2x^4-{}^{6}C_3x^3
\displaystyle \qquad\qquad +{}^{6}C_4x^2-{}^{6}C_5x+{}^{6}C_6.
\displaystyle =x^6-6x^5+15x^4-20x^3+15x^2-6x+1.
\displaystyle \therefore (x^2+1-2x)^3=x^6-6x^5+15x^4-20x^3+15x^2-6x+1.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Using binomial theorem, evaluate: }(0.99)^5.
\displaystyle \text{Answer:}
\displaystyle (0.99)^5=(1-0.01)^5.
\displaystyle \text{By the Binomial Theorem,}
\displaystyle (1-0.01)^5=1-{}^5C_1(0.01)+{}^5C_2(0.01)^2-{}^5C_3(0.01)^3
\displaystyle \qquad\qquad +{}^5C_4(0.01)^4-{}^5C_5(0.01)^5.
\displaystyle =1-5(0.01)+10(0.0001)-10(0.000001)+5(0.00000001)-0.0000000001.
\displaystyle =1-0.05+0.001-0.00001+0.00000005-0.0000000001.
\displaystyle =0.9509900499.
\displaystyle \therefore (0.99)^5=0.9509900499 = 0.951
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Find the coefficient of }x^5\text{ in the expansion of }\left(3x^2+\frac{1}{3x}\right)^{10}.
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}={}^{10}C_r(3x^2)^{10-r}\left(\frac{1}{3x}\right)^r.
\displaystyle ={}^{10}C_r\,3^{10-r}x^{20-2r}\cdot3^{-r}x^{-r}.
\displaystyle ={}^{10}C_r\,3^{10-2r}x^{20-3r}.
\displaystyle \text{For the term containing }x^5,
\displaystyle 20-3r=5.
\displaystyle \therefore 3r=15\Rightarrow r=5.
\displaystyle \therefore \text{Coefficient of }x^5={} ^{10}C_5\,3^{10-2(5)}.
\displaystyle ={}^{10}C_5=252.
\displaystyle \therefore \text{The coefficient of }x^5\text{ is }252.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Write general term in the expansion of }\left(3z^2-\frac{3}{3z^2}\right)^{35}.
\displaystyle \text{Answer:}
\displaystyle \text{For }(a+b)^n,\text{ the general term is}
\displaystyle T_{r+1}={}^nC_r a^{n-r}b^r.
\displaystyle \text{Here, }a=3z^2,\quad b=-\frac{3}{3z^2},\quad n=35.
\displaystyle \therefore T_{r+1}={}^{35}C_r(3z^2)^{35-r}\left(-\frac{3}{3z^2}\right)^r.
\displaystyle =(-1)^r{}^{35}C_r(3z^2)^{35-r}\left(\frac{1}{z^2}\right)^r.
\displaystyle =(-1)^r{}^{35}C_r3^{35-r}z^{70-4r}.
\displaystyle \therefore T_{r+1}=(-1)^r{}^{35}C_r3^{35-r}z^{70-4r},\quad 0\leq r\leq35.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Using binomial theorem, show that }3\text{ is a factor of }7^n-4^n,\text{ where }n\in N.
\displaystyle \text{Answer:}
\displaystyle 7^n-4^n=(4+3)^n-4^n.
\displaystyle \text{By the Binomial Theorem,}
\displaystyle (4+3)^n=4^n+{}^nC_1 4^{n-1}(3)+{}^nC_2 4^{n-2}(3)^2+\cdots+3^n.
\displaystyle \therefore 7^n-4^n={}^nC_1 4^{n-1}(3)+{}^nC_2 4^{n-2}(3)^2+\cdots+3^n.
\displaystyle =3\left[{}^nC_1 4^{n-1}+{}^nC_2 4^{n-2}(3)+\cdots+3^{n-1}\right].
\displaystyle \therefore 7^n-4^n\text{ is divisible by }3.
\displaystyle \therefore 3\text{ is a factor of }7^n-4^n.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Show that sum of powers of }a\text{ and }b\text{ of each term in the expansion of}
\displaystyle (x+a)^n\text{ is always }n.
\displaystyle \text{Answer:}
\displaystyle \text{By the Binomial Theorem, the general term in the expansion of }(x+a)^n\text{ is}
\displaystyle T_{r+1}={}^nC_r x^{n-r}a^r.
\displaystyle \text{In this term, the power of }x\text{ is }n-r\text{ and the power of }a\text{ is }r.
\displaystyle \therefore \text{sum of the powers}=(n-r)+r=n.
\displaystyle \therefore \text{the sum of the powers of the variables in every term is always }n.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \text{SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 31: }\text{If the coefficients of }x^7\text{ and }x^8\text{ in the expansion of }\left(2+\frac{x}{3}\right)^n
\displaystyle \text{are equal, then find }n.
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion of }\left(2+\frac{x}{3}\right)^n\text{ is}
\displaystyle T_{r+1}={}^nC_r2^{n-r}\left(\frac{x}{3}\right)^r.
\displaystyle ={}^nC_r2^{n-r}\frac{x^r}{3^r}.
\displaystyle \therefore \text{Coefficient of }x^7={}^nC_7\frac{2^{n-7}}{3^7}.
\displaystyle \text{Coefficient of }x^8={}^nC_8\frac{2^{n-8}}{3^8}.
\displaystyle \text{Since the coefficients are equal,}
\displaystyle {}^nC_7\frac{2^{n-7}}{3^7}={}^nC_8\frac{2^{n-8}}{3^8}.
\displaystyle \frac{{}^nC_8}{{}^nC_7}=6.
\displaystyle \frac{n-7}{8}=6.
\displaystyle n-7=48.
\displaystyle \therefore n=55.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Find the term independent of }x\text{ in the expansion of}
\displaystyle \left(\sqrt[3]{x}+\frac{1}{2\sqrt[3]{x}}\right)^{18},\quad x>0.
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}={}^{18}C_r\left(x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{1/3}}\right)^r.
\displaystyle ={}^{18}C_r\frac{1}{2^r}x^{\frac{18-r}{3}-\frac{r}{3}}.
\displaystyle ={}^{18}C_r\frac{1}{2^r}x^{\frac{18-2r}{3}}.
\displaystyle \text{For the term independent of }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle \frac{18-2r}{3}=0.
\displaystyle 18-2r=0.
\displaystyle \therefore r=9.
\displaystyle \therefore \text{the required term is }T_{10}={}^{18}C_9\frac{1}{2^9}.
\displaystyle =\frac{48620}{512}=\frac{12155}{128}.
\displaystyle \therefore \text{The term independent of }x\text{ is }\frac{12155}{128}.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If the coefficient of }(r-5)^{th}\text{ and }(2r-1)^{th}\text{ terms in the expansion of}
\displaystyle (1+x)^{34}\text{ are equal, find }r.
\displaystyle \text{Answer:}
\displaystyle \text{The coefficient of the }k^{th}\text{ term in }(1+x)^{34}\text{ is }{}^{34}C_{k-1}.
\displaystyle \therefore \text{coefficient of }(r-5)^{th}\text{ term}={}^{34}C_{r-6}.
\displaystyle \text{Coefficient of }(2r-1)^{th}\text{ term}={}^{34}C_{2r-2}.
\displaystyle \text{Given, }{}^{34}C_{r-6}={}^{34}C_{2r-2}.
\displaystyle \text{For equal binomial coefficients, either the suffixes are equal or their sum is }34.
\displaystyle \text{If }r-6=2r-2,\text{ then }r=-4,\text{ which is not admissible.}
\displaystyle \therefore (r-6)+(2r-2)=34.
\displaystyle 3r-8=34.
\displaystyle 3r=42.
\displaystyle \therefore r=14.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Find the coefficient of }a^4\text{ in the product }(1+2a)^4(2-a)^5\text{ using}
\displaystyle \text{binomial theorem.}
\displaystyle \text{Answer:}
\displaystyle (1+2a)^4=\sum_{r=0}^{4}{}^4C_r(2a)^r.
\displaystyle (2-a)^5=\sum_{s=0}^{5}{}^5C_s2^{5-s}(-a)^s.
\displaystyle \text{For the coefficient of }a^4,\text{ we require }r+s=4.
\displaystyle \therefore \text{Coefficient of }a^4
\displaystyle ={}^4C_0{}^5C_4(2)^1-{}^4C_1(2){}^5C_3(2)^2
\displaystyle \qquad+{}^4C_2(2)^2{}^5C_2(2)^3-{}^4C_3(2)^3{}^5C_1(2)^4
\displaystyle \qquad+{}^4C_4(2)^4{}^5C_0(2)^5.
\displaystyle =10-320+1920-2560+512.
\displaystyle =-438.
\displaystyle \therefore \text{The coefficient of }a^4\text{ is }-438.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{The sum of the coefficients of the first three terms in the expansion}
\displaystyle \text{of }\left(x-\frac{3}{x^2}\right)^m,\ x\ne0,\ m\text{ being a natural number, is }559.\text{ Find the term}
\displaystyle \text{of the expansion containing }x^3.
\displaystyle \text{Answer:}
\displaystyle \text{The general term is}
\displaystyle T_{r+1}={}^mC_r x^{m-r}\left(-\frac{3}{x^2}\right)^r.
\displaystyle ={}^mC_r(-3)^r x^{m-3r}.
\displaystyle \text{The coefficients of the first three terms are}
\displaystyle 1,\quad -3m,\quad 9\,{}^mC_2.
\displaystyle \text{Given, }1-3m+9\,{}^mC_2=559.
\displaystyle 1-3m+\frac{9m(m-1)}{2}=559.
\displaystyle 9m^2-15m-1116=0.
\displaystyle 3m^2-5m-372=0.
\displaystyle (3m+31)(m-12)=0.
\displaystyle \text{Since }m\text{ is a natural number, }m=12.
\displaystyle \text{Now, }T_{r+1}={}^{12}C_r(-3)^r x^{12-3r}.
\displaystyle \text{For the term containing }x^3,
\displaystyle 12-3r=3.
\displaystyle \therefore r=3.
\displaystyle \therefore T_4={}^{12}C_3(-3)^3x^3.
\displaystyle =220(-27)x^3=-5940x^3.
\displaystyle \therefore \text{The term containing }x^3\text{ is }-5940x^3.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Show that the coefficient of the middle term in the expansion of}
\displaystyle (1+x)^{2n}\text{ is equal to the sum of the coefficients of two middle terms}
\displaystyle \text{in the expansion of }(1+x)^{2n-1}.
\displaystyle \text{Answer:}
\displaystyle \text{The expansion of }(1+x)^{2n}\text{ contains }2n+1\text{ terms.}
\displaystyle \therefore \text{its middle term is the }(n+1)^{th}\text{ term.}
\displaystyle \text{Coefficient of the middle term}={}^{2n}C_n.
\displaystyle \text{The expansion of }(1+x)^{2n-1}\text{ contains }2n\text{ terms.}
\displaystyle \therefore \text{its two middle terms are the }n^{th}\text{ and }(n+1)^{th}\text{ terms.}
\displaystyle \text{Their coefficients are }{}^{2n-1}C_{n-1}\text{ and }{}^{2n-1}C_n.
\displaystyle \text{By Pascal's identity,}
\displaystyle {}^{2n-1}C_{n-1}+{}^{2n-1}C_n={} ^{2n}C_n.
\displaystyle \therefore \text{the coefficient of the middle term of }(1+x)^{2n}\text{ equals the sum}
\displaystyle \text{of the coefficients of the two middle terms of }(1+x)^{2n-1}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If }p\text{ is a real number and the middle term in the expansion of }\left(p+\frac{2}{2}\right)^8
\displaystyle \text{is }1120,\text{ then find the value of }p.
\displaystyle \text{Answer:}
\displaystyle \left(p+\frac{2}{2}\right)^8=(p+1)^8.
\displaystyle \text{Since the expansion has }9\text{ terms, the middle term is the }5^{th}\text{ term.}
\displaystyle T_5={}^8C_4p^{8-4}(1)^4.
\displaystyle =70p^4.
\displaystyle \text{Given, }70p^4=1120.
\displaystyle \therefore p^4=16.
\displaystyle \therefore p=\pm2.
\displaystyle \text{Hence, the values of }p\text{ are }2\text{ and }-2.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Find the coefficient of }x^4\text{ in the expansion of }(1+x+x^2+x^3)^{11}.
\displaystyle \text{Answer:}
\displaystyle 1+x+x^2+x^3=\frac{1-x^4}{1-x}.
\displaystyle \therefore (1+x+x^2+x^3)^{11}=(1-x^4)^{11}(1-x)^{-11}.
\displaystyle \text{Now, }(1-x^4)^{11}=1-11x^4+\text{terms of degree higher than }4.
\displaystyle \text{Also, }(1-x)^{-11}=\sum_{r=0}^{\infty}{}^{10+r}C_{10}x^r.
\displaystyle \text{Coefficient of }x^4\text{ in }(1-x)^{-11}={}^{14}C_{10}={}^{14}C_4=1001.
\displaystyle \text{Therefore, coefficient of }x^4
\displaystyle =1001-11=990.
\displaystyle \therefore \text{The coefficient of }x^4\text{ is }990.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If the coefficients of }2^{nd},\ 3^{rd}\text{ and }4^{th}\text{ terms in the expansion of }
\displaystyle (1+x)^{2n} \ \text{are in A.P., show that }2n^2-9n+7=0.
\displaystyle \text{Answer:}
\displaystyle \text{The coefficients of the }2^{nd},\ 3^{rd}\text{ and }4^{th}\text{ terms are}
\displaystyle {}^{2n}C_1,\qquad{}^{2n}C_2,\qquad{}^{2n}C_3.
\displaystyle \text{Since these coefficients are in A.P.,}
\displaystyle 2\,{}^{2n}C_2={} ^{2n}C_1+{}^{2n}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{Multiplying throughout by }6,\text{ we get}
\displaystyle 12n(2n-1)=12n+2n(2n-1)(2n-2).
\displaystyle \text{Dividing by }2n,\text{ we get}
\displaystyle 6(2n-1)=6+(2n-1)(2n-2).
\displaystyle 12n-6=6+4n^2-6n+2.
\displaystyle 4n^2-18n+14=0.
\displaystyle \therefore 2n^2-9n+7=0.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Find the }6^{th}\text{ term of the expansion }\left(y^{1/2}+x^{1/3}\right)^n,\text{ if the coefficient of }3^{rd}
\displaystyle \text{term from end is }45.
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion is}
\displaystyle T_{r+1}={}^nC_r\left(y^{1/2}\right)^{n-r}\left(x^{1/3}\right)^r.
\displaystyle \text{The }3^{rd}\text{ term from the end has coefficient }{}^nC_{n-2}={}^nC_2.
\displaystyle \text{Given, }{}^nC_2=45.
\displaystyle \frac{n(n-1)}{2}=45.
\displaystyle n(n-1)=90.
\displaystyle n^2-n-90=0.
\displaystyle (n-10)(n+9)=0.
\displaystyle \text{Since }n\text{ is positive, }n=10.
\displaystyle \text{The }6^{th}\text{ term is obtained by putting }r=5.
\displaystyle T_6={} ^{10}C_5\left(y^{1/2}\right)^5\left(x^{1/3}\right)^5.
\displaystyle =252y^{5/2}x^{5/3}.
\displaystyle \therefore \text{The }6^{th}\text{ term is }252x^{5/3}y^{5/2}.
\displaystyle \\

\displaystyle \text{LONG ANSWER QUESTIONS }

\displaystyle \textbf{Question 41: }\text{Find the term independent of }x\text{ in the expansion of}
\displaystyle (1+x+2x^3)\left(\frac{3x^2}{2}-\frac{1}{3x}\right).
\displaystyle \text{Answer:}
\displaystyle (1+x+2x^3)\left(\frac{3x^2}{2}-\frac{1}{3x}\right)
\displaystyle =\frac{3x^2}{2}+\frac{3x^3}{2}+3x^5-\frac{1}{3x}-\frac{1}{3}-\frac{2x^2}{3}.
\displaystyle \text{The only term independent of }x\text{ is }-\frac{1}{3}.
\displaystyle \therefore \text{The term independent of }x\text{ is }-\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Find }n,\text{ if ratio of the }7^{th}\text{ term from beginning and }7^{th}\text{ term from end}
\displaystyle \text{in the expansion of }\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n\text{ is }\frac{1}{6}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a=\sqrt[3]{2}\text{ and }b=\frac{1}{\sqrt[3]{3}}.
\displaystyle \text{The }7^{th}\text{ term from the beginning is}
\displaystyle T_7={}^nC_6a^{n-6}b^6.
\displaystyle \text{The }7^{th}\text{ term from the end is}
\displaystyle T_{n-5}={}^nC_6a^6b^{n-6}.
\displaystyle \text{Given, }\frac{T_7}{T_{n-5}}=\frac{1}{6}.
\displaystyle \therefore \frac{a^{n-6}b^6}{a^6b^{n-6}}=\frac{1}{6}.
\displaystyle \therefore \left(\frac{a}{b}\right)^{n-12}=\frac{1}{6}.
\displaystyle \frac{a}{b}=\sqrt[3]{2}\times\sqrt[3]{3}=\sqrt[3]{6}.
\displaystyle \therefore \left(\sqrt[3]{6}\right)^{n-12}=\frac{1}{6}.
\displaystyle 6^{\frac{n-12}{3}}=6^{-1}.
\displaystyle \therefore \frac{n-12}{3}=-1.
\displaystyle n-12=-3.
\displaystyle \therefore n=9.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{The }2^{nd},\ 3^{rd}\text{ and }4^{th}\text{ terms in the expansion of }(x+a)^n\text{ are }240,\ 720
\displaystyle \text{and }1080\text{ respectively. Find }x,\ a\text{ and }n.
\displaystyle \text{Answer:}
\displaystyle \text{The general term in the expansion of }(x+a)^n\text{ is}
\displaystyle T_{r+1}={}^nC_r x^{n-r}a^r.
\displaystyle \therefore T_2=nx^{n-1}a=240.\qquad ...(1)
\displaystyle T_3={}^nC_2x^{n-2}a^2=720.\qquad ...(2)
\displaystyle T_4={}^nC_3x^{n-3}a^3=1080.\qquad ...(3)

\displaystyle \text{Dividing (2) by (1), we get}
\displaystyle \frac{n-1}{2}\cdot\frac{a}{x}=\frac{720}{240}=3.
\displaystyle \therefore \frac{a}{x}=\frac{6}{n-1}.\qquad ...(4)

\displaystyle \text{Dividing (3) by (2), we get}
\displaystyle \frac{n-2}{3}\cdot\frac{a}{x}=\frac{1080}{720}=\frac{3}{2}.
\displaystyle \therefore \frac{a}{x}=\frac{9}{2(n-2)}.\qquad ...(5)

\displaystyle \text{From (4) and (5),}
\displaystyle \frac{6}{n-1}=\frac{9}{2(n-2)}.
\displaystyle 12(n-2)=9(n-1).
\displaystyle 12n-24=9n-9.
\displaystyle \therefore 3n=15\Rightarrow n=5.

\displaystyle \text{From (4), }\frac{a}{x}=\frac{6}{4}=\frac{3}{2}.
\displaystyle \therefore a=\frac{3x}{2}.

\displaystyle \text{Using (1), }5x^4a=240.
\displaystyle x^4a=48.
\displaystyle x^4\left(\frac{3x}{2}\right)=48.
\displaystyle \frac{3}{2}x^5=48.
\displaystyle x^5=32.
\displaystyle \therefore x=2.
\displaystyle \therefore a=\frac{3}{2}(2)=3.
\displaystyle \therefore x=2,\qquad a=3,\qquad n=5.
\displaystyle \\

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