JEE Advanced Maths PYQ: Binomial Theorem (Chapterwise Questions & Solutions)

Q 1 :

For $r = 0, 1, \dots, 10,$ let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of ${(1 + x)}^{10}, {(1 + x)}^{20}$ and ${(1 + x)}^{30}$ . Then $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to [2010]

$B_{10} - C_{10}$
$A_{10} ({B^{2}}_{10} C_{10} A_{10})$
$0$
$C_{10} - B_{10}$

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(4)

$Clearly A_{r} = {}^{10}C_{r}, B_{r} = {}^{20}C_{r}, C_{r} = {}^{30}C_{r}$

Now $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$

$= \sum_{r = 1}^{10} {}^{10}C_{r} ({}^{20}C_{10} {}^{20}C_{r} - {}^{30}C_{10} {}^{30}C_{r})$

$= {}^{20}C_{10} \sum_{r = 1}^{10} {}^{10}C_{r} {}^{20}C_{r} - {}^{30}C_{10} \sum_{r = 1}^{10} {}^{10}C_{r} \cdot {}^{10}C_{r}$

$= {}^{20}C_{10} ({}^{10}C_{1} {}^{20}C_{1} + {}^{10}C_{2} {}^{20}C_{2} + \dots + {}^{10}C_{10} {}^{20}C_{10})$ $- {}^{30}C_{10} ({}^{10}C_{1} \times {}^{10}C_{1} + {}^{10}C_{2} \times {}^{10}C_{2} + \dots + {}^{10}C_{10} \times {}^{10}C_{10}) ...(i)$

Now, on expanding ${(1 + x)}^{10}$ and ${(1 + x)}^{20}$ and comparing coefficients of $x^{20}$ in their product on both sides, we get

${}^{10}C_{0} {}^{20}C_{0} + {}^{10}C_{1} {}^{20}C_{1} + {}^{10}C_{2} {}^{20}C_{2} + \dots + {}^{10}C_{10} {}^{20}C_{10}$

$= Coeff. of x^{20} in {(1 + x)}^{30}$ $= {}^{30}C_{20} = {}^{30}C_{10}$

$∴$ ${}^{10}C_{1} {}^{20}C_{1} + {}^{10}C_{2} {}^{20}C_{2} + \dots + {}^{10}C_{10} {}^{20}C_{10} = {}^{30}C_{10} - 1$ $. . . (i i)$

$Again on expanding {(1 + x)}^{10} and {(x + 1)}^{10} and comparing the coefficients of x^{10} in their product on both sides, we get$

$∴$ ${({}^{10}C_{0})}^{2} + {({}^{10}C_{1})}^{2} + \dots + {({}^{10}C_{10})}^{2}$

$= Coeff. of x^{10} in {(1 + x)}^{20}$ $= {}^{20}C_{10}$

$∴$ ${({}^{10}C_{1})}^{2} + {({}^{10}C_{2})}^{2} + \dots + {({}^{10}C_{10})}^{2} = {}^{20}C_{10} - 1$ $. . . (i i i)$

Now from equations (i), (ii), and (iii), we get

$Required value = {}^{20}C_{10} ({}^{30}C_{10} - 1) - {}^{30}C_{10} ({}^{20}C_{10} - 1)$

$= {}^{30}C_{10} - {}^{20}C_{10} = C_{10} - B_{10}$

Q 2 :

Coefficient of $t^{24}$ in ${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$ is [2003]

${}^{12}C_{6} + 3$
${}^{12}C_{6} + 1$
${}^{12}C_{6}$
${}^{12}C_{6} + 2$

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(4)

${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$

$= (1 + t^{12} + t^{24} + t^{36}) {(1 + t^{2})}^{12}$

$∴$ $Coeff. of t^{24} = 1 \times Coeff. of t^{24} in {(1 + t^{2})}^{12} + 1 \times Coeff. of t^{12} in {(1 + t^{2})}^{12} + 1 \times constant term in {(1 + t^{2})}^{12}$

$= {}^{12}C_{12} + {}^{12}C_{6} + {}^{12}C_{0} = 1 + {}^{12}C_{6} + 1 = {}^{12}C_{6} + 2$

Q 3 :

In the binomial expansion of ${(a - b)}^{n}, n \geq 5,$ the sum of the 5th and 6th terms is zero. Then $\frac{a}{b}$ equals [2001]

$\frac{(n - 5)}{6}$
$\frac{(n - 4)}{5}$
$\frac{5}{(n - 4)}$
$\frac{6}{(n - 5)}$

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(2)

$In binomial expansion {(a - b)}^{n}, n \geq 5;$

$T_{5} + T_{6} = 0$

$\Rightarrow {}^{n}C_{4} a^{n - 4} b^{4} - {}^{n}C_{5} a^{n - 5} b^{5} = 0$

$\Rightarrow \frac{{}^{n}C_{4}}{{}^{n}C_{5}} \cdot \frac{a}{b} = 1$

$\Rightarrow \frac{5}{n - 4} \cdot \frac{a}{b} = 1$

$\Rightarrow \frac{a}{b} = \frac{n - 4}{5}$

Q 4 :

Let $a$ and $b$ be two non-zero real numbers. If the coefficient of $x^{5}$ in the expansion of ${(a x^{2} + \frac{70}{27 b x})}^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{7},$ then the value of $2 b$ is [2023]

(3)

$Given expansion {(a x^{2} + \frac{70}{27 b x})}^{4}$

$T_{r + 1} = {}^{4}C_{r} {(a x^{2})}^{4 - r} {(\frac{70}{27 b x})}^{r}$

$= {}^{4}C_{r} a^{4 - r} {(\frac{70}{27 b})}^{r} . x^{8 - 3 r}$

$Here, 8 - 3 r = 5 \Rightarrow r = 1$

$So, coefficient of x^{5} = {}^{4}C_{1} a^{3} \cdot \frac{70}{27 b}$

$For expansion {(a x - \frac{1}{b x^{2}})}^{7}$

$T_{r + 1} = {}^{7}C_{r} {(a x)}^{7 - r} {(\frac{- 1}{b x^{2}})}^{r} = {}^{7}C_{r} a^{7 - r} {(\frac{- 1}{b})}^{r} x^{7 - 3 r}$

$Here, 7 - 3 r = - 5 \Rightarrow r = 4$

$So, coefficient of x^{- 5} = {}^{7}C_{4} a^{3} {(\frac{- 1}{b})}^{3}$

$A.T.Q., {}^{4}C_{1} a^{3} \cdot \frac{70}{27 b} = {}^{7}C_{4} a^{3} \cdot \frac{- 1}{b^{3}}$

$\Rightarrow b = \frac{3}{2} \Rightarrow 2 b = 3$

Q 5 :

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of ${(1 + x)}^{2} + {(1 + x)}^{3} + \dots + {(1 + x)}^{49} + {(1 + m x)}^{50}$ is $(3 n + 1) {}^{51}C_{3}$ for some positive integer $n$ . Then the value of $n$ is [2016]

(5)

${(1 + x)}^{2} + {(1 + x)}^{3} + \dots + {(1 + x)}^{49} + {(1 + m x)}^{50}$

$= {(1 + x)}^{2} [\frac{{(1 + x)}^{48} - 1}{(1 + x) - 1}] + {(1 + m x)}^{50}$

$= \frac{1}{x} [{(1 + x)}^{50} - {(1 + x)}^{2}] + {(1 + m x)}^{50}$

$Coeff. of x^{2} in the above expansion$

$= Coeff. of x^{3} in {(1 + x)}^{50} + Coeff. of x^{2} in {(1 + m x)}^{50}$

$= {}^{50}C_{3} + {}^{50}C_{2} m^{2}$

$∴$ $(3 n + 1) {}^{51}C_{3} = {}^{50}C_{3} + {}^{50}C_{2} m^{2}$

$\Rightarrow (3 n + 1) = \frac{{}^{50}C_{3}}{{}^{51}C_{3}} + \frac{{}^{50}C_{2}}{{}^{51}C_{3}} m^{2}$

$\Rightarrow 3 n + 1 = \frac{16}{17} + \frac{1}{17} m^{2}$

$\Rightarrow n = \frac{m^{2} - 1}{51}$

$∴$ Least positive integer $m$ for which $n$ is an integer is $m = 16$ and then $n = 5$

Q 6 :

The coefficients of three consecutive terms of ${(1 + x)}^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n =$ [2013]

(6)

Let the coefficients of three consecutive terms of ${(1 + x)}^{n + 5}$ be ${}^{n + 5}C_{r - 1}, {}^{n + 5}C_{r}, {}^{n + 5}C_{r + 1}$ ,

then we have ${}^{n + 5}C_{r - 1} : {}^{n + 5}C_{r} : {}^{n + 5}C_{r + 1} = 5 : 10 : 14$

$\frac{{}^{n + 5}C_{r - 1}}{{}^{n + 5}C_{r}} = \frac{5}{10} \Rightarrow \frac{r}{n + 6 - r} = \frac{1}{2}$

$\Rightarrow n - 3 r + 6 = 0 . . . (i)$

$Also \frac{{}^{n + 5}C_{r}}{{}^{n + 5}C_{r + 1}} = \frac{10}{14} \Rightarrow \frac{r + 1}{n - r + 5} = \frac{5}{7}$

$\Rightarrow 5 n - 12 r + 18 = 0 . . . (i i)$

$Solving (i) and (ii), we get n = 6$

Q 7 :

Let $X = {({}^{10}{C_{1}})}^{2} + 2 {({}^{10}{C_{2}})}^{2} + 3 {({}^{10}{C_{3}})}^{2} + \dots + 10 {({}^{10}{C_{10}})}^{2},$ where ${}^{10}{C_{r}}, r \in {1, 2, \dots, 10}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is ______ . [2018]

(646)

$\sum_{r = 0}^{n} r {({}^{n}C_{r})}^{2} = n \sum_{r = 0}^{n} {}^{n}C_{r} {}^{n - 1}C_{r - 1}$

$= n \sum_{r = 1}^{n} {}^{n}C_{n - r} {}^{n - 1}C_{r - 1} = n {}^{2 n - 1}C_{n - 1}$

$Now, X = {({}^{10}C_{1})}^{2} + 2 {({}^{10}C_{2})}^{2} + 3 {({}^{10}C_{3})}^{2} + \dots + 10 {({}^{10}C_{10})}^{2}$

$= \sum_{n = 0}^{10} r {({}^{10}C_{r})}^{2} = 10 {}^{19}C_{9}$

$∴$ $\frac{X}{1430} = \frac{1}{143} {}^{19}C_{9} = 646$

Topic Question Set

Q 1 :

For $r = 0, 1, \dots, 10,$ let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of ${(1 + x)}^{10}, {(1 + x)}^{20}$ and ${(1 + x)}^{30}$ . Then $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to [2010]

Q 2 :

Coefficient of $t^{24}$ in ${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$ is [2003]

Q 3 :

In the binomial expansion of ${(a - b)}^{n}, n \geq 5,$ the sum of the 5th and 6th terms is zero. Then $\frac{a}{b}$ equals [2001]

Q 4 :

Let $a$ and $b$ be two non-zero real numbers. If the coefficient of $x^{5}$ in the expansion of ${(a x^{2} + \frac{70}{27 b x})}^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{7},$ then the value of $2 b$ is [2023]

Q 5 :

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of ${(1 + x)}^{2} + {(1 + x)}^{3} + \dots + {(1 + x)}^{49} + {(1 + m x)}^{50}$ is $(3 n + 1) {}^{51}C_{3}$ for some positive integer $n$ . Then the value of $n$ is [2016]

Q 6 :

The coefficients of three consecutive terms of ${(1 + x)}^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n =$ [2013]

Q 7 :

Let $X = {({}^{10}{C_{1}})}^{2} + 2 {({}^{10}{C_{2}})}^{2} + 3 {({}^{10}{C_{3}})}^{2} + \dots + 10 {({}^{10}{C_{10}})}^{2},$ where ${}^{10}{C_{r}}, r \in {1, 2, \dots, 10}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is ______ . [2018]

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Topic Question Set

Q 1 : For r=0,1,…,10, let Ar,Br and Cr denote, respectively, the coefficient of xr in the expansions of (1+x)10,(1+x)20 and (1+x)30. Then ∑r=110Ar(B10Br-C10Ar) is equal to [2010]

Q 2 : Coefficient of t24 in (1+t2)12(1+t12)(1+t24) is [2003]

Q 3 : In the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero. Then ab equals [2001]

Q 4 : Let a and b be two non-zero real numbers. If the coefficient of x5 in the expansion of (ax2+7027bx)4 is equal to the coefficient of x-5 in the expansion of (ax-1bx2)7, then the value of 2b is [2023]

Q 5 : Let m be the smallest positive integer such that the coefficient of x2 in the expansion of (1+x)2+(1+x)3+⋯+(1+x)49+(1+mx)50 is (3n+1) C351 for some positive integer n. Then the value of n is [2016]

Q 6 : The coefficients of three consecutive terms of (1+x)n+5 are in the ratio 5:10:14. Then n= [2013]

Q 7 : Let X=(C110)2+2(C210)2+3(C310)2+⋯+10(C1010)2, where Cr10, r∈{1,2,…,10} denote binomial coefficients. Then, the value of 11430X is ______ . [2018]

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Q 1 :

For $r = 0, 1, \dots, 10,$ let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of ${(1 + x)}^{10}, {(1 + x)}^{20}$ and ${(1 + x)}^{30}$ . Then $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to [2010]

Q 2 :

Coefficient of $t^{24}$ in ${(1 + t^{2})}^{12} (1 + t^{12}) (1 + t^{24})$ is [2003]

Q 3 :

In the binomial expansion of ${(a - b)}^{n}, n \geq 5,$ the sum of the 5th and 6th terms is zero. Then $\frac{a}{b}$ equals [2001]

Q 4 :

Let $a$ and $b$ be two non-zero real numbers. If the coefficient of $x^{5}$ in the expansion of ${(a x^{2} + \frac{70}{27 b x})}^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of ${(a x - \frac{1}{b x^{2}})}^{7},$ then the value of $2 b$ is [2023]

Q 5 :

Let $m$ be the smallest positive integer such that the coefficient of $x^{2}$ in the expansion of ${(1 + x)}^{2} + {(1 + x)}^{3} + \dots + {(1 + x)}^{49} + {(1 + m x)}^{50}$ is $(3 n + 1) {}^{51}C_{3}$ for some positive integer $n$ . Then the value of $n$ is [2016]

Q 6 :

The coefficients of three consecutive terms of ${(1 + x)}^{n + 5}$ are in the ratio $5 : 10 : 14$ . Then $n =$ [2013]

Q 7 :

Let $X = {({}^{10}{C_{1}})}^{2} + 2 {({}^{10}{C_{2}})}^{2} + 3 {({}^{10}{C_{3}})}^{2} + \dots + 10 {({}^{10}{C_{10}})}^{2},$ where ${}^{10}{C_{r}}, r \in {1, 2, \dots, 10}$ denote binomial coefficients. Then, the value of $\frac{1}{1430} X$ is ______ . [2018]