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

\displaystyle \textbf{Question 1: }\text{If in the expansion of }(1+x)^{20}, \text{ the coefficients of }r\text{th and }(r+4)\text{th terms are equal, then }r\text{ is equal to}
\displaystyle \text{(a) }7\qquad \text{(b) }8\qquad \text{(c) }9\qquad \text{(d) }10
\displaystyle \text{Answer: }
\displaystyle \text{The coefficient of the }r\text{th term is }{}^{20}C_{r-1}.
\displaystyle \text{The coefficient of the }(r+4)\text{th term is }{}^{20}C_{r+3}.
\displaystyle \text{Since the coefficients are equal,}
\displaystyle {}^{20}C_{r-1}={}^{20}C_{r+3}.
\displaystyle \text{Using }{}^nC_x={}^nC_y,\text{ where }x\ne y,\text{ we have }x+y=n.
\displaystyle (r-1)+(r+3)=20.
\displaystyle 2r+2=20.
\displaystyle 2r=18\Rightarrow r=9.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The term without }x\text{ in the expansion of } \left(2x-\frac{1}{2x^2}\right)^{12}\text{ is}
\displaystyle \text{(a) }495\qquad \text{(b) }-495\qquad \text{(c) }-7920\qquad \text{(d) }7920
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{12}C_r(2x)^{12-r} \left(-\frac{1}{2x^2}\right)^r.
\displaystyle T_{r+1}={}^{12}C_r(-1)^r2^{12-2r}x^{12-3r}.
\displaystyle \text{For the term without }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle 12-3r=0\Rightarrow r=4.
\displaystyle \therefore T_5={}^{12}C_4(-1)^4 2^{12-8}.
\displaystyle =495\times16=7920.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }r\text{th term in the expansion of } \left(2x^2-\frac{1}{x}\right)^{12}\text{ is without }x,\text{ then }r\text{ is equal to}
\displaystyle \text{(a) }8\qquad \text{(b) }7\qquad \text{(c) }9\qquad \text{(d) }10
\displaystyle \text{Answer: }
\displaystyle \text{The }r\text{th term is}
\displaystyle T_r={}^{12}C_{r-1}(2x^2)^{12-(r-1)} \left(-\frac{1}{x}\right)^{r-1}.
\displaystyle T_r={}^{12}C_{r-1}2^{13-r}(-1)^{r-1} x^{\,2(13-r)-(r-1)}.
\displaystyle \text{For the term without }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle 2(13-r)-(r-1)=0.
\displaystyle 26-2r-r+1=0.
\displaystyle 27-3r=0\Rightarrow r=9.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If in the expansion of }(a+b)^n\text{ and }(a+b)^{n+3},
\displaystyle \text{the ratio of the coefficients of second and third terms, and third and fourth}
\displaystyle \text{terms respectively are equal, then }n\text{ is}
\displaystyle \text{(a) }3\qquad \text{(b) }4\qquad \text{(c) }5\qquad \text{(d) }6
\displaystyle \text{Answer: }
\displaystyle \text{In }(a+b)^n,\text{ the coefficients of the second and third terms are}
\displaystyle {}^nC_1\text{ and }{}^nC_2\text{ respectively.}
\displaystyle \therefore \frac{{}^nC_1}{{}^nC_2} =\frac{n}{\frac{n(n-1)}{2}}=\frac{2}{n-1}.
\displaystyle \text{In }(a+b)^{n+3},\text{ the coefficients of the third and fourth terms are}
\displaystyle {}^{n+3}C_2\text{ and }{}^{n+3}C_3\text{ respectively.}
\displaystyle \therefore \frac{{}^{n+3}C_2}{{}^{n+3}C_3} =\frac{3}{n+1}.
\displaystyle \text{Since the given ratios are equal,}
\displaystyle \frac{2}{n-1}=\frac{3}{n+1}.
\displaystyle 2(n+1)=3(n-1).
\displaystyle 2n+2=3n-3\Rightarrow n=5.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }A\text{ and }B\text{ are the sums of odd and even terms respectively}
\displaystyle \text{in the expansion of }(x+a)^n,\text{ then }(x+a)^{2n}-(x-a)^{2n}\text{ is equal to}
\displaystyle \text{(a) }4(A+B)\qquad \text{(b) }4(A-B)\qquad \text{(c) }AB\qquad \text{(d) }4AB
\displaystyle \text{Answer: }
\displaystyle \text{Since }A\text{ and }B\text{ are the sums of odd and even numbered terms,}
\displaystyle A+B=(x+a)^n.
\displaystyle A-B=(x-a)^n.
\displaystyle \therefore (x+a)^{2n}-(x-a)^{2n}=(A+B)^2-(A-B)^2.
\displaystyle =(A+B+A-B)(A+B-A+B).
\displaystyle =(2A)(2B)=4AB.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The number of irrational terms in the expansion of } \left(4^{1/5}+7^{1/10}\right)^{45}\text{ is}
\displaystyle \text{(a) }40\qquad \text{(b) }5\qquad \text{(c) }41\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{45}C_r \left(4^{1/5}\right)^{45-r}\left(7^{1/10}\right)^r.
\displaystyle T_{r+1}={}^{45}C_r 4^{(45-r)/5}7^{r/10}.
\displaystyle \text{For the term to be rational, }\frac{45-r}{5} \text{ and }\frac{r}{10}\text{ must be integers.}
\displaystyle \frac{r}{10}\text{ is an integer when } r=0,10,20,30,40.
\displaystyle \text{For each of these values, }\frac{45-r}{5} \text{ is also an integer.}
\displaystyle \therefore \text{Number of rational terms}=5.
\displaystyle \text{Total number of terms}=45+1=46.
\displaystyle \therefore \text{Number of irrational terms}=46-5=41.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The coefficient of }x^{-17} \text{ in the expansion of }\left(x^4-\frac{1}{x^3}\right)^{15}\text{ is}
\displaystyle \text{(a) }1365\qquad \text{(b) }-1365\qquad \text{(c) }3003\qquad \text{(d) }-3003
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{15}C_r(x^4)^{15-r} \left(-\frac{1}{x^3}\right)^r.
\displaystyle T_{r+1}=(-1)^r{}^{15}C_r x^{60-7r}.
\displaystyle \text{For the term containing }x^{-17},
\displaystyle 60-7r=-17.
\displaystyle 7r=77\Rightarrow r=11.
\displaystyle \therefore \text{Coefficient of }x^{-17} =(-1)^{11}{}^{15}C_{11}.
\displaystyle =-{}^{15}C_4=-1365.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the expansion of } \left(x^2-\frac{1}{3x}\right)^9,\text{ the term without }x\text{ is equal to}
\displaystyle \text{(a) }\frac{28}{81}\qquad \text{(b) }-\frac{28}{243}\qquad \text{(c) }\frac{28}{243}\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{9}C_r(x^2)^{9-r} \left(-\frac{1}{3x}\right)^r.
\displaystyle T_{r+1}={}^{9}C_r(-1)^r \frac{x^{18-3r}}{3^r}.
\displaystyle \text{For the term without }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle 18-3r=0\Rightarrow r=6.
\displaystyle \therefore T_7={}^{9}C_6(-1)^6\frac{1}{3^6}.
\displaystyle =\frac{84}{729}=\frac{28}{243}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If in the expansion of }(1+x)^{15}, \text{ the coefficients of }(2r+3)\text{th}
\displaystyle \text{and }(r-1)\text{th terms are equal, then the value of }r\text{ is}
\displaystyle \text{(a) }5\qquad \text{(b) }6\qquad \text{(c) }4\qquad \text{(d) }3
\displaystyle \text{Answer: }
\displaystyle \text{Coefficient of the }(2r+3)\text{th term} ={}^{15}C_{2r+2}.
\displaystyle \text{Coefficient of the }(r-1)\text{th term} ={}^{15}C_{r-2}.
\displaystyle \text{Since the coefficients are equal,}
\displaystyle {}^{15}C_{2r+2}={}^{15}C_{r-2}.
\displaystyle \text{Using }{}^nC_p={}^nC_q,\text{ where }p\ne q, \text{ we have }p+q=n.
\displaystyle (2r+2)+(r-2)=15.
\displaystyle 3r=15\Rightarrow r=5.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The middle term in the expansion of } \left(\frac{2x^2}{3}+\frac{3}{2x^2}\right)^{10}\text{ is}
\displaystyle \text{(a) }251\qquad \text{(b) }252\qquad \text{(c) }250\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle \text{Since }n=10,\text{ the number of terms is }10+1=11.
\displaystyle \therefore \text{The middle term is the }6\text{th term.}
\displaystyle T_{r+1}={}^{10}C_r \left(\frac{2x^2}{3}\right)^{10-r} \left(\frac{3}{2x^2}\right)^r.
\displaystyle \therefore T_6={}^{10}C_5 \left(\frac{2x^2}{3}\right)^5 \left(\frac{3}{2x^2}\right)^5.
\displaystyle T_6={}^{10}C_5=252.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If in the expansion of } \left(x^4-\frac{1}{x^3}\right)^{15},\ x^{-17}\text{ occurs in }r\text{th term, then}
\displaystyle \text{(a) }r=10\qquad \text{(b) }r=11\qquad \text{(c) }r=12\qquad \text{(d) }r=13
\displaystyle \text{Answer: }
\displaystyle T_{r}={}^{15}C_{r-1}(x^4)^{15-(r-1)} \left(-\frac{1}{x^3}\right)^{r-1}.
\displaystyle T_r={}^{15}C_{r-1}(-1)^{r-1} x^{4(16-r)-3(r-1)}.
\displaystyle \text{Since the term contains }x^{-17},
\displaystyle 4(16-r)-3(r-1)=-17.
\displaystyle 64-4r-3r+3=-17.
\displaystyle 67-7r=-17.
\displaystyle 7r=84\Rightarrow r=12.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the expansion of } \left(x-\frac{1}{3x^2}\right)^9,\text{ the term independent of }x\text{ is}
\displaystyle \text{(a) }T_3\qquad \text{(b) }T_4\qquad \text{(c) }T_5\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{9}C_r x^{9-r} \left(-\frac{1}{3x^2}\right)^r.
\displaystyle T_{r+1}={}^{9}C_r(-1)^r \frac{x^{9-3r}}{3^r}.
\displaystyle \text{For the term independent of }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle 9-3r=0.
\displaystyle r=3.
\displaystyle \therefore \text{The required term is }T_{r+1}=T_4.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If in the expansion of }(1+y)^n, \text{ the coefficients of 5th, 6th and 7th terms are in A.P.,}
\displaystyle \text{then }n\text{ is equal to}
\displaystyle \text{(a) }7,11\qquad \text{(b) }7,14\qquad \text{(c) }8,16\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle \text{The coefficients of the 5th, 6th and 7th terms are } {}^nC_4,\ {}^nC_5,\ {}^nC_6.
\displaystyle \text{Since these coefficients are in A.P.,}
\displaystyle 2{}^nC_5={}^nC_4+{}^nC_6.
\displaystyle \text{Dividing throughout by }{}^nC_5,
\displaystyle 2=\frac{{}^nC_4}{{}^nC_5} +\frac{{}^nC_6}{{}^nC_5}.
\displaystyle 2=\frac{5}{n-4}+\frac{n-5}{6}.
\displaystyle 12(n-4)=30+(n-5)(n-4).
\displaystyle 12n-48=30+n^2-9n+20.
\displaystyle n^2-21n+98=0.
\displaystyle (n-7)(n-14)=0.
\displaystyle \therefore n=7\text{ or }14.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In the expansion of } \left(\frac{1}{2}x^{1/3}+x^{-1/5}\right)^8,\text{ the term independent of }x\text{ is}
\displaystyle \text{(a) }T_5\qquad \text{(b) }T_6\qquad \text{(c) }T_7\qquad \text{(d) }T_8
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{8}C_r \left(\frac{1}{2}x^{1/3}\right)^{8-r} \left(x^{-1/5}\right)^r.
\displaystyle T_{r+1}={}^{8}C_r\left(\frac{1}{2}\right)^{8-r} x^{\frac{8-r}{3}-\frac{r}{5}}.
\displaystyle \text{For the term independent of }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle \frac{8-r}{3}-\frac{r}{5}=0.
\displaystyle 5(8-r)-3r=0.
\displaystyle 40-8r=0\Rightarrow r=5.
\displaystyle \therefore \text{The required term is }T_{r+1}=T_6.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If the sum of odd numbered terms and the sum of even numbered}
\displaystyle \text{terms in the expansion of }(x+a)^n\text{ are }A\text{ and }B\text{ respectively, then the value}
\displaystyle \text{of }(x^2-a^2)^n\text{ is}
\displaystyle \text{(a) }A^2-B^2\qquad \text{(b) }A^2+B^2\qquad \text{(c) }4AB\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle \text{Since }A\text{ and }B\text{ are the sums of odd and even numbered terms,}
\displaystyle A+B=(x+a)^n.
\displaystyle A-B=(x-a)^n.
\displaystyle \therefore (A+B)(A-B)=(x+a)^n(x-a)^n.
\displaystyle A^2-B^2=\{(x+a)(x-a)\}^n.
\displaystyle A^2-B^2=(x^2-a^2)^n.
\displaystyle \therefore (x^2-a^2)^n=A^2-B^2.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If the coefficient of }x\text{ in } \left(x^2+\frac{\lambda}{x}\right)^5\text{ is }270,\text{ then }\lambda=
\displaystyle \text{(a) }3\qquad \text{(b) }4\qquad \text{(c) }5\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{5}C_r(x^2)^{5-r} \left(\frac{\lambda}{x}\right)^r.
\displaystyle T_{r+1}={}^{5}C_r\lambda^r x^{10-3r}.
\displaystyle \text{For the term containing }x,\text{ we must have}
\displaystyle 10-3r=1.
\displaystyle 3r=9\Rightarrow r=3.
\displaystyle \therefore \text{Coefficient of }x={}^{5}C_3\lambda^3.
\displaystyle \text{Given that the coefficient of }x\text{ is }270,
\displaystyle {}^{5}C_3\lambda^3=270.
\displaystyle 10\lambda^3=270\Rightarrow\lambda^3=27.
\displaystyle \therefore \lambda=3.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The coefficient of }x^4\text{ in } \left(\frac{x}{2}-\frac{3}{x^2}\right)^{10}\text{ is}
\displaystyle \text{(a) }\frac{405}{256}\qquad \text{(b) }\frac{504}{259}\qquad \text{(c) }\frac{450}{263}\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{10}C_r \left(\frac{x}{2}\right)^{10-r} \left(-\frac{3}{x^2}\right)^r.
\displaystyle T_{r+1}={}^{10}C_r \frac{(-3)^r}{2^{10-r}}x^{10-3r}.
\displaystyle \text{For the term containing }x^4,\text{ we must have}
\displaystyle 10-3r=4.
\displaystyle 3r=6\Rightarrow r=2.
\displaystyle \therefore \text{Coefficient of }x^4 ={}^{10}C_2\frac{(-3)^2}{2^8}.
\displaystyle =45\times\frac{9}{256} =\frac{405}{256}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The total number of terms in the expansion of } \\ (x+a)^{100}+(x-a)^{100}\text{ after simplification is}
\displaystyle \text{(a) }202\qquad \text{(b) }51\qquad \text{(c) }50\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle (x+a)^{100}+(x-a)^{100}
\displaystyle =2\left\{{}^{100}C_0x^{100} +{}^{100}C_2x^{98}a^2+\cdots+{}^{100}C_{100}a^{100}\right\}.
\displaystyle \text{Thus, only the terms containing even powers of }a\text{ remain.}
\displaystyle \text{These correspond to }r=0,2,4,\ldots,100.
\displaystyle \therefore \text{Number of terms} =\frac{100}{2}+1=51.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }T_2/T_3\text{ in the expansion of }(a+b)^n \text{ and }T_3/T_4\text{ in the expansion of}
\displaystyle (a+b)^{n+3}\text{ are equal, then }n=
\displaystyle \text{(a) }3\qquad \text{(b) }4\qquad \text{(c) }5\qquad \text{(d) }6
\displaystyle \text{Answer: }
\displaystyle \text{In }(a+b)^n,\quad T_2={}^nC_1a^{n-1}b,\quad T_3={}^nC_2a^{n-2}b^2.
\displaystyle \therefore \frac{T_2}{T_3} =\frac{{}^nC_1}{{}^nC_2}\frac{a}{b} =\frac{2}{n-1}\frac{a}{b}.
\displaystyle \text{In }(a+b)^{n+3},\quad T_3={}^{n+3}C_2a^{n+1}b^2,\quad T_4={}^{n+3}C_3a^nb^3.
\displaystyle \therefore \frac{T_3}{T_4} =\frac{{}^{n+3}C_2}{{}^{n+3}C_3}\frac{a}{b} =\frac{3}{n+1}\frac{a}{b}.
\displaystyle \text{Since the given ratios are equal,}
\displaystyle \frac{2}{n-1}\frac{a}{b} =\frac{3}{n+1}\frac{a}{b}.
\displaystyle \frac{2}{n-1}=\frac{3}{n+1}.
\displaystyle 2(n+1)=3(n-1).
\displaystyle 2n+2=3n-3\Rightarrow n=5.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The coefficient of }\frac{1}{x}\text{ in the expansion of } (1+x)^n\left(1+\frac{1}{x}\right)^n\text{ is}
\displaystyle \text{(a) }\frac{n!}{(n-1)!(n+1)!}\qquad \text{(b) }\frac{(2n)!}{(n-1)!(n+1)!}
\displaystyle \text{(c) }\frac{(2n)!}{(2n-1)!(2n+1)!}\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle (1+x)^n\left(1+\frac{1}{x}\right)^n =(1+x)^n\left(\frac{1+x}{x}\right)^n.
\displaystyle =\frac{(1+x)^{2n}}{x^n}.
\displaystyle \text{For the resulting term to contain }\frac{1}{x}, \text{ we require the term containing }x^{n-1}\text{ in }(1+x)^{2n}.
\displaystyle \text{Coefficient of }x^{n-1} ={}^{2n}C_{n-1}.
\displaystyle {}^{2n}C_{n-1} =\frac{(2n)!}{(n-1)!\{2n-(n-1)\}!}.
\displaystyle =\frac{(2n)!}{(n-1)!(n+1)!}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If the sum of the binomial coefficients of the expansion } \left(2x+\frac{1}{x}\right)^n
\displaystyle \text{ is equal to }256,\text{ then the term independent} \ \text{of }x\text{ is}
\displaystyle \text{(a) }1120\qquad \text{(b) }1020\qquad \text{(c) }512\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle \text{The sum of the binomial coefficients is }2^n.
\displaystyle 2^n=256=2^8.
\displaystyle \therefore n=8.
\displaystyle \text{Now, }T_{r+1}={}^{8}C_r(2x)^{8-r} \left(\frac{1}{x}\right)^r.
\displaystyle T_{r+1}={}^{8}C_r2^{8-r}x^{8-2r}.
\displaystyle \text{For the term independent of }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle 8-2r=0\Rightarrow r=4.
\displaystyle \therefore T_5={}^{8}C_4 2^{8-4}.
\displaystyle =70\times16=1120.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If the fifth term of the expansion } \left(a^{2/3}+a^{-1}\right)^n\text{ does not contain }a,\text{ then }n\text{ is equal to}
\displaystyle \text{(a) }2\qquad \text{(b) }5\qquad \text{(c) }10\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^nC_r\left(a^{2/3}\right)^{n-r} \left(a^{-1}\right)^r.
\displaystyle \text{For the fifth term, }r+1=5\Rightarrow r=4.
\displaystyle \therefore T_5={}^nC_4 a^{\frac{2}{3}(n-4)}a^{-4}.
\displaystyle ={}^nC_4a^{\frac{2n-8}{3}-4}.
\displaystyle ={}^nC_4a^{\frac{2n-20}{3}}.
\displaystyle \text{Since the fifth term does not contain }a,\text{ its exponent must be zero.}
\displaystyle \frac{2n-20}{3}=0.
\displaystyle 2n-20=0\Rightarrow n=10.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The coefficient of }x^{-3}\text{ in the expansion of } \left(x-\frac{m}{x}\right)^{11}\text{ is}
\displaystyle \text{(a) }-924m^7\qquad \text{(b) }-792m^5\qquad \text{(c) }-792m^6\qquad \text{(d) }-330m^7
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{11}C_r x^{11-r} \left(-\frac{m}{x}\right)^r.
\displaystyle T_{r+1}={}^{11}C_r(-1)^r m^r x^{11-2r}.
\displaystyle \text{For the term containing }x^{-3},
\displaystyle 11-2r=-3.
\displaystyle 2r=14\Rightarrow r=7.
\displaystyle \therefore \text{Coefficient of }x^{-3} ={}^{11}C_7(-1)^7m^7.
\displaystyle =-{}^{11}C_7m^7.
\displaystyle =-{}^{11}C_4m^7.
\displaystyle =-330m^7.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{The coefficient of the term independent of }x \text{ in the expansion of }\left(ax+\frac{b}{x}\right)^{14}\text{ is}
\displaystyle \text{(a) }14!\,a^7b^7\qquad \text{(b) }\frac{14!}{7!}a^7b^7\qquad \text{(c) }\frac{14!}{(7!)^2}a^7b^7\qquad \text{(d) }\frac{14!}{(7!)^3}a^7b^7
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{14}C_r(ax)^{14-r} \left(\frac{b}{x}\right)^r.
\displaystyle T_{r+1}={}^{14}C_r a^{14-r}b^r x^{14-2r}.
\displaystyle \text{For the term independent of }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle 14-2r=0\Rightarrow r=7.
\displaystyle \therefore \text{Required coefficient} ={}^{14}C_7a^7b^7.
\displaystyle =\frac{14!}{7!7!}a^7b^7 =\frac{14!}{(7!)^2}a^7b^7.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{The coefficient of }x^5\text{ in the expansion of } \\ (1+x)^{21}+(1+x)^{22}+\cdots+(1+x)^{30}\text{ is}
\displaystyle \text{(a) }{}^{51}C_5\qquad \text{(b) }{}^9C_5\qquad \text{(c) }{}^{31}C_6-{}^{21}C_6\qquad \text{(d) }{}^{30}C_5+{}^{20}C_5
\displaystyle \text{Answer: }
\displaystyle \text{Coefficient of }x^5 ={}^{21}C_5+{}^{22}C_5+\cdots+{}^{30}C_5.
\displaystyle \text{Using }{}^nC_r+{}^nC_{r+1}={}^{n+1}C_{r+1},
\displaystyle \sum_{k=21}^{30}{}^kC_5 ={}^{31}C_6-{}^{21}C_6.
\displaystyle \therefore \text{The coefficient of }x^5 ={}^{31}C_6-{}^{21}C_6.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The coefficient of }x^8y^{10}\text{ in the expansion of } (x+y)^{18}\text{ is}
\displaystyle \text{(a) }{}^{18}C_8\qquad \text{(b) }{}^{18}P_{10}\qquad \text{(c) }2^{18}\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle T_{r+1}={}^{18}C_r x^{18-r}y^r.
\displaystyle \text{For the term }x^8y^{10},\text{ we have }r=10.
\displaystyle \therefore \text{Coefficient of }x^8y^{10}={}^{18}C_{10}.
\displaystyle {}^{18}C_{10}={}^{18}C_{18-10}={}^{18}C_8.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If the coefficients of the }(n+1)^{th}\text{ term and the } (n+3)^{th}
\displaystyle \text{ term in the expansion of }(1+x)^{20}\text{ are equal, then the}  \ \text{value of }n\text{ is}
\displaystyle \text{(a) }10\qquad \text{(b) }8\qquad \text{(c) }9\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle \text{Coefficient of the }(n+1)^{th}\text{ term is }{}^{20}C_n.
\displaystyle \text{Coefficient of the }(n+3)^{th}\text{ term is }{}^{20}C_{n+2}.
\displaystyle \text{Since the coefficients are equal,}
\displaystyle {}^{20}C_n={}^{20}C_{n+2}.
\displaystyle \text{Using }{}^NC_r={}^NC_s\Rightarrow r=s\text{ or }r+s=N,
\displaystyle n+(n+2)=20.
\displaystyle 2n+2=20\Rightarrow 2n=18\Rightarrow n=9.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If the coefficients of 2nd, 3rd and 4th terms in the expansion of }
\displaystyle (1+x)^n,\ n\in N \ \text{are in A.P., then }n=
\displaystyle \text{(a) }7\qquad \text{(b) }14\qquad \text{(c) }2\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle \text{The coefficients of the 2nd, 3rd and 4th terms are } {}^nC_1,\ {}^nC_2,\ {}^nC_3.
\displaystyle \text{Since they 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 n-1=1+\frac{(n-1)(n-2)}{6}.
\displaystyle 6n-12=(n-1)(n-2).
\displaystyle 6n-12=n^2-3n+2.
\displaystyle n^2-9n+14=0.
\displaystyle (n-7)(n-2)=0.
\displaystyle \therefore n=7\text{ or }n=2.
\displaystyle \text{But for the 4th term to exist, }n\geq3.
\displaystyle \therefore n=7.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{The middle term in the expansion of } \left(\frac{2x}{3}-\frac{3}{2x^2}\right)^{2n}\text{ is}
\displaystyle \text{(a) }{}^{2n}C_n\qquad \text{(b) }(-1)^n\,{}^{2n}C_nx^{-n}\qquad \text{(c) }{}^{2n}C_nx^{-n}\qquad \text{(d) none of these}
\displaystyle \text{Answer: }
\displaystyle \text{The expansion has }2n+1\text{ terms, so the middle term is }T_{n+1}.
\displaystyle T_{n+1}={}^{2n}C_n \left(\frac{2x}{3}\right)^{2n-n} \left(-\frac{3}{2x^2}\right)^n.
\displaystyle ={}^{2n}C_n \left(\frac{2x}{3}\right)^n \left(-\frac{3}{2x^2}\right)^n.
\displaystyle ={}^{2n}C_n(-1)^n \left(\frac{2x}{3}\cdot\frac{3}{2x^2}\right)^n.
\displaystyle =(-1)^n\,{}^{2n}C_n\left(\frac{1}{x}\right)^n.
\displaystyle =(-1)^n\,{}^{2n}C_nx^{-n}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }r^{th}\text{ term is the middle term in the expansion of } \left(x^2-\frac{1}{2x}\right)^{20}, \\ \text{ then }(r+3)^{th}\text{ term is}
\displaystyle \text{(a) }{}^{20}C_{14}\left(\frac{x}{2^{14}}\right)\qquad \text{(b) }{}^{20}C_{12}x^2\,2^{-12}\qquad \text{(c) }-{}^{20}C_7x\,2^{-13}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The expansion has }20+1=21\text{ terms.}
\displaystyle \therefore \text{The middle term is the }11^{th}\text{ term.}
\displaystyle \therefore r=11.
\displaystyle \therefore r+3=14.
\displaystyle \text{Thus, we have to find }T_{14}.
\displaystyle T_{14}=T_{13+1}={}^{20}C_{13} (x^2)^{20-13}\left(-\frac{1}{2x}\right)^{13}.
\displaystyle ={}^{20}C_{13}x^{14} \frac{(-1)^{13}}{2^{13}x^{13}}.
\displaystyle =-{}^{20}C_{13}x\,2^{-13}.
\displaystyle \text{Since }{}^{20}C_{13}={}^{20}C_7,
\displaystyle T_{14}=-{}^{20}C_7x\,2^{-13}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{The number of terms with integral coefficients in the expansion of } \\ \left(17^{1/3}+35^{1/2}x\right)^{600}\text{ is}
\displaystyle \text{(a) }100\qquad \text{(b) }50\qquad \text{(c) }150\qquad \text{(d) }101
\displaystyle \text{Answer:}
\displaystyle T_{r+1}={}^{600}C_r \left(17^{1/3}\right)^{600-r} \left(35^{1/2}x\right)^r.
\displaystyle T_{r+1}={}^{600}C_r 17^{(600-r)/3}35^{r/2}x^r.
\displaystyle \text{For the coefficient to be integral, } \frac{600-r}{3}\text{ and }\frac{r}{2}\text{ must be integers.}
\displaystyle \frac{600-r}{3}\in\mathbb Z \Rightarrow r\text{ is divisible by }3.
\displaystyle \frac{r}{2}\in\mathbb Z \Rightarrow r\text{ is divisible by }2.
\displaystyle \therefore r\text{ must be divisible by }6.
\displaystyle r=0,6,12,\ldots,600.
\displaystyle \text{Number of possible values of }r =\frac{600}{6}+1=101.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Constant term in the expansion of } \left(x-\frac{1}{x}\right)^{10}\text{ is}
\displaystyle \text{(a) }152\qquad \text{(b) }-152\qquad \text{(c) }-252\qquad \text{(d) }252
\displaystyle \text{Answer:}
\displaystyle T_{r+1}={}^{10}C_r x^{10-r} \left(-\frac{1}{x}\right)^r.
\displaystyle T_{r+1}=(-1)^r{}^{10}C_r x^{10-2r}.
\displaystyle \text{For the constant term, the exponent of }x\text{ must be zero.}
\displaystyle 10-2r=0\Rightarrow r=5.
\displaystyle \therefore T_6=(-1)^5{}^{10}C_5.
\displaystyle =-{}^{10}C_5=-252.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If the coefficients of }x^2\text{ and }x^3 \text{ in the expansion of }(3+ax)^9 \\ \text{are the same, then the value of }a\text{ is}
\displaystyle \text{(a) }-\frac79\qquad \text{(b) }-\frac97\qquad \text{(c) }\frac79\qquad \text{(d) }\frac97
\displaystyle \text{Answer:}
\displaystyle T_{r+1}={}^9C_r3^{9-r}(ax)^r.
\displaystyle \text{Coefficient of }x^2 ={}^9C_2\,3^7a^2.
\displaystyle \text{Coefficient of }x^3 ={}^9C_3\,3^6a^3.
\displaystyle \text{Since the coefficients are equal,}
\displaystyle {}^9C_2\,3^7a^2 ={}^9C_3\,3^6a^3.
\displaystyle {}^9C_2\cdot3 ={}^9C_3a.
\displaystyle a=3\frac{{}^9C_2}{{}^9C_3}.
\displaystyle a=3\left(\frac{36}{84}\right) =3\left(\frac37\right)=\frac97.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the number of terms in the expansion of } \\ (2+\sqrt{3x})^{10}+(2-\sqrt{3x})^{10}.
\displaystyle \text{Answer:}
\displaystyle \text{On adding the two expansions, all terms containing odd powers of } \sqrt{3x}\text{ cancel.}
\displaystyle \text{Only the terms corresponding to }r=0,2,4,6,8,10\text{ remain.}
\displaystyle \therefore \text{Number of terms}=6.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the sum of the coefficients in the expansion of } \\ (1-3x+x^2)^{111}.
\displaystyle \text{Answer:}
\displaystyle \text{The sum of the coefficients is obtained by putting }x=1.
\displaystyle (1-3+1)^{111}=(-1)^{111}=-1.
\displaystyle \therefore \text{The sum of the coefficients is }-1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Write the number of terms in the expansion of } \\ (1-3x+3x^2-x^3)^8.
\displaystyle \text{Answer:}
\displaystyle 1-3x+3x^2-x^3=(1-x)^3.
\displaystyle \therefore (1-3x+3x^2-x^3)^8=(1-x)^{24}.
\displaystyle \text{The expansion of }(1-x)^{24}\text{ contains }24+1=25\text{ terms.}
\displaystyle \therefore \text{The number of terms is }25.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Write the middle term in the expansion of } \left(\frac{2x^2}{3}+\frac{3}{2x^2}\right)^{10}.
\displaystyle \text{Answer:}
\displaystyle \text{There are }10+1=11\text{ terms, so the middle term is }T_6.
\displaystyle T_6={}^{10}C_5 \left(\frac{2x^2}{3}\right)^5 \left(\frac{3}{2x^2}\right)^5.
\displaystyle ={}^{10}C_5 \left(\frac{2x^2}{3}\cdot\frac{3}{2x^2}\right)^5.
\displaystyle ={}^{10}C_5=252.
\displaystyle \therefore \text{The middle term is }252.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Which term is independent of }x\text{ in the expansion of } \left(x-\frac{1}{3x^2}\right)^9?
\displaystyle \text{Answer:}
\displaystyle T_{r+1}={}^9C_r x^{9-r} \left(-\frac{1}{3x^2}\right)^r.
\displaystyle T_{r+1}={}^9C_r(-1)^r3^{-r}x^{9-3r}.
\displaystyle \text{For the term independent of }x,\text{ the exponent of }x\text{ must be zero.}
\displaystyle 9-3r=0\Rightarrow r=3.
\displaystyle \therefore r+1=4.
\displaystyle \therefore \text{The }4^{th}\text{ term }T_4\text{ is independent of }x.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }a\text{ and }b\text{ denote respectively the coefficients of }x^m \text{ and }x^n
\displaystyle \text{ in the expansion of }(1+x)^{m+n},\text{ then write the} \ \text{relation between }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle a={}^{m+n}C_m.
\displaystyle b={}^{m+n}C_n.
\displaystyle \text{Using }{}^NC_r={}^NC_{N-r},
\displaystyle {}^{m+n}C_m={}^{m+n}C_{m+n-m}={}^{m+n}C_n.
\displaystyle \therefore a=b.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }a\text{ and }b\text{ are coefficients of }x^n \text{ in the expansions of }(1+x)^{2n}\text{ and }
\displaystyle (1+x)^{2n-1}  \ \text{respectively, then write the relation between }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle a={}^{2n}C_n.
\displaystyle b={}^{2n-1}C_n.
\displaystyle \therefore \frac{a}{b} =\frac{{}^{2n}C_n}{{}^{2n-1}C_n}.
\displaystyle =\frac{\dfrac{(2n)!}{n!n!}} {\dfrac{(2n-1)!}{n!(n-1)!}}.
\displaystyle =\frac{(2n)!}{(2n-1)!}\cdot \frac{(n-1)!}{n!}.
\displaystyle =\frac{2n}{n}=2.
\displaystyle \therefore a=2b.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the middle term in the expansion of } \left(x+\frac{1}{x}\right)^{10}.
\displaystyle \text{Answer:}
\displaystyle \text{The expansion contains }10+1=11\text{ terms.}
\displaystyle \therefore \text{The middle term is the }6^{th}\text{ term, i.e. }T_6.
\displaystyle T_6={}^{10}C_5x^{10-5} \left(\frac{1}{x}\right)^5.
\displaystyle ={}^{10}C_5x^5\cdot x^{-5}.
\displaystyle ={}^{10}C_5=252.
\displaystyle \therefore \text{The middle term is }252.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }a\text{ and }b\text{ denote the sum of the coefficients in the expansions of}
\displaystyle (1-3x+10x^2)^n\text{ and }(1+x^2)^n\text{ respectively, then write the relation between }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{The sum of the coefficients is obtained by putting }x=1.
\displaystyle a=(1-3+10)^n=8^n.
\displaystyle b=(1+1)^n=2^n.
\displaystyle a=8^n=(2^3)^n=(2^n)^3.
\displaystyle \therefore a=b^3.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write the coefficient of the middle term in the expansion of } \\ (1+x)^{2n}.
\displaystyle \text{Answer:}
\displaystyle \text{The expansion of }(1+x)^{2n}\text{ contains }2n+1\text{ terms.}
\displaystyle \therefore \text{The middle term is the }(n+1)^{th}\text{ term.}
\displaystyle T_{n+1}={}^{2n}C_n(1)^{2n-n}x^n.
\displaystyle ={}^{2n}C_nx^n.
\displaystyle \therefore \text{The coefficient of the middle term is }{}^{2n}C_n.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the number of terms in the expansion of } \left\{(2x+y^3)^4\right\}^7.
\displaystyle \text{Answer:}
\displaystyle \left\{(2x+y^3)^4\right\}^7=(2x+y^3)^{28}.
\displaystyle \text{The expansion of }(a+b)^n\text{ contains }n+1\text{ terms.}
\displaystyle \therefore \text{Number of terms}=28+1=29.
\displaystyle \therefore \text{The number of terms is }29.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the sum of the coefficients of two middle terms in the} \\ \text{binomial expansion of }(1+x)^{2n-1}.
\displaystyle \text{Answer:}
\displaystyle \text{Since the index }2n-1\text{ is odd, there are two middle terms.}
\displaystyle \text{They 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{Using }{}^NC_r={}^NC_{N-r},
\displaystyle {}^{2n-1}C_{n-1}={}^{2n-1}C_n.
\displaystyle \therefore \text{Sum}=2\,{}^{2n-1}C_{n-1}.
\displaystyle \text{Also, by Pascal's identity,}
\displaystyle {}^{2n-1}C_{n-1}+{}^{2n-1}C_n={}^{2n}C_n.
\displaystyle \therefore \text{The required sum is }{}^{2n}C_n.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the ratio of the coefficients of }x^p\text{ and }x^q \text{ in the expansion of } \\ (1+x)^{p+q}.
\displaystyle \text{Answer:}
\displaystyle \text{Coefficient of }x^p={}^{p+q}C_p.
\displaystyle \text{Coefficient of }x^q={}^{p+q}C_q.
\displaystyle \text{Using }{}^NC_r={}^NC_{N-r},
\displaystyle {}^{p+q}C_p={}^{p+q}C_{p+q-p}={}^{p+q}C_q.
\displaystyle \therefore \text{Required ratio}=1:1.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Write last two digits of the number }3^{400}.
\displaystyle \text{Answer:}
\displaystyle 3^{20}=3486784401.
\displaystyle \therefore 3^{20}\equiv 1\pmod{100}.
\displaystyle 3^{400}=(3^{20})^{20}.
\displaystyle \therefore 3^{400}\equiv 1^{20}\equiv 1\pmod{100}.
\displaystyle \therefore \text{The last two digits are }01.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the number of terms in the expansion of }(a+b+c)^n.
\displaystyle \text{Answer:}
\displaystyle \text{A general term is of the form }a^p b^q c^r,\text{ where}
\displaystyle p+q+r=n,\qquad p,q,r\geq0.
\displaystyle \text{Thus, the number of terms equals the number of non-negative integral solutions of}
\displaystyle p+q+r=n.
\displaystyle \therefore \text{Number of terms}={}^{n+2}C_2.
\displaystyle =\frac{(n+2)(n+1)}{2}.
\displaystyle \therefore \text{The number of terms is }\frac{(n+1)(n+2)}{2}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }a\text{ and }b\text{ are the coefficients of }x^n \text{ in the expansions of } \\ (1+x)^{2n}\text{ and }(1+x)^{2n-1}\text{ respectively, find }\frac{a}{b}.
\displaystyle \text{Answer:}
\displaystyle a={}^{2n}C_n,\qquad b={}^{2n-1}C_n.
\displaystyle \therefore \frac{a}{b} =\frac{{}^{2n}C_n}{{}^{2n-1}C_n}.
\displaystyle =\frac{\dfrac{(2n)!}{n!n!}} {\dfrac{(2n-1)!}{n!(n-1)!}}.
\displaystyle =\frac{(2n)!}{(2n-1)!}\cdot\frac{(n-1)!}{n!}.
\displaystyle =\frac{2n}{n}=2.
\displaystyle \therefore \frac{a}{b}=2.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Write the total number of terms in the expansion of } \\ (x+a)^{100}+(x-a)^{100}.
\displaystyle \text{Answer:}
\displaystyle (x+a)^{100}+(x-a)^{100}
\displaystyle =2\left\{{}^{100}C_0x^{100} +{}^{100}C_2x^{98}a^2+\cdots+{}^{100}C_{100}a^{100}\right\}.
\displaystyle \text{The terms containing odd powers of }a\text{ cancel.}
\displaystyle \text{Thus, the remaining terms correspond to }r=0,2,4,\ldots,100.
\displaystyle \therefore \text{Number of terms}=\frac{100}{2}+1=51.
\displaystyle \therefore \text{The total number of terms is }51.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }(1-x+x^2)^n =a_0+a_1x+a_2x^2+\cdots+a_{2n}x^{2n},\text{ find the value of}
\displaystyle a_0+a_2+a_4+\cdots+a_{2n}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x)=(1-x+x^2)^n.
\displaystyle P(1)=a_0+a_1+a_2+\cdots+a_{2n}=1^n=1.
\displaystyle P(-1)=a_0-a_1+a_2-a_3+\cdots+a_{2n}=3^n.
\displaystyle \text{Adding these two equations,}
\displaystyle P(1)+P(-1)=2(a_0+a_2+a_4+\cdots+a_{2n}).
\displaystyle 1+3^n=2(a_0+a_2+a_4+\cdots+a_{2n}).
\displaystyle \therefore a_0+a_2+a_4+\cdots+a_{2n} =\frac{3^n+1}{2}.
\displaystyle \\

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