Binomial Theorem and its Simple Applications jee mains questions | Binomial Theorem and its Simple Applications Previous Year Question

Q 31 :
If the second, third and fourth terms in the expansion of ${(x + y)}^{n}$ are 135, 30, and $\frac{10}{3},$ respectively, then $6 (n^{3} + x^{2} + y)$ is equal to ______. [2024]

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(806)

$T_{r + 1} = {}^{n}C_{r} x^{n - r} y^{r}$

$∴$ ${}^{n}{C_{1}} x^{n - 1} y = 135,$ ...(i)

${}^{n}{C_{2}} x^{n - 2} y^{2} = 30$ ...(ii)

and ${}^{n}{C_{3}} x^{n - 3} y^{3} = \frac{10}{3}$ ...(iii)

By (i) and (ii), we have

$\frac{{}^{n}{C_{1}} x}{{}^{n}{C_{2}} y} = \frac{135}{30} = \frac{9}{2}$ ...(iv)

By (ii) and (iii), we have

$\frac{{}^{n}{C_{2}} x}{{}^{n}{C_{3}} y} = \frac{30}{\frac{10}{3}} = 9$ ...(v)

By (iv) and (v), we have

$\frac{{}^{n}{C_{1}} x}{{}^{n}{C_{2}} y} \times \frac{{}^{n}{C_{3}} y}{{}^{n}{C_{2}} x} = \frac{\frac{9}{2}}{9}$

$\Rightarrow \frac{{}^{n}{C_{1} {}^{n}{C_{3}}}}{{({}^{n}{C_{2}})}^{2}} = \frac{1}{2} \Rightarrow 2 {}^{n}{C_{1}} {}^{n}{C_{3}} = {({}^{n}{C_{2}})}^{2}$

$\Rightarrow \frac{2 n n!}{3! (n - 3)!} = {(\frac{n!}{2! (n - 2)!})}^{2}$

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

$\Rightarrow \frac{n^{2} (n - 1) (n - 2)}{3} = \frac{n^{2} {(n - 1)}^{2}}{4} \Rightarrow \frac{n - 2}{3} = \frac{n - 1}{4}$

$\Rightarrow 4 n - 8 = 3 n - 3 \Rightarrow n = 5$

$\Rightarrow \frac{{}^{5}{C_{2} x}}{{}^{5}{C_{3}} y} = 9$ [Using (v)]

$\Rightarrow \frac{x}{y} = 9$

$\Rightarrow x = 9 y$ ...(vi)

$∴$ From (i), ${}^{5}{C_{1}} x^{4} \frac{x}{9} = 135 \Rightarrow x^{5} = \frac{135 \times 9}{5}$

$\Rightarrow x^{5} = 3^{5}$

$\Rightarrow x = 3$ and $y = \frac{1}{3}$ [Using (vi)]

Hence, $6 (n^{3} + x^{2} + y) = 6 (125 + 9 + \frac{1}{3}) = 806$

Q 32 :
The coefficient of $x^{2012}$ in the expansion of ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$ is equal to _______ . [2024]

0%

(0)

We have, ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$

$= {(1 - x)}^{2007} (1 - x) {(1 + x + x^{2})}^{2007} = (1 - x) {(1 - x^{3})}^{2007}$

$\Rightarrow (1 - x) ({}^{2007}{C_{0} - {}^{2007}{C_{1}} (x^{3}) + . . .})$

General term is $(1 - x) ({(- 1)}^{r} {}^{2007}{C_{r}} x^{3 r})$

$= {(- 1)}^{r} {}^{2007}{C_{r}} x^{3 r} - {(- 1)}^{r} {}^{2007}{C_{r}} x^{3 r + 1}$

Now, $3 r = 2012 \Rightarrow r \neq \frac{2012}{3}$

So, $3 r + 1 = 2012 \Rightarrow 3 r = 2011 \Rightarrow r \neq \frac{2011}{3}$

Hence, there is no term containing $x^{2012} .$

So, coefficient of $x^{2012} = 0$

Q 33 :
Number of integral terms in the expansion of ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$ is equal to _______ . [2024]

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(138)

We have, ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$

Now, ${(n + 1)}^{th}$ term $= t_{n + 1} = {}^{n}{C_{r}} 11^{\frac{r}{6}} 7^{\frac{824 - r}{2}}$

For integral term, 6 should divide $r$ and $\frac{824 - r}{2}$ must be integer.

$\Rightarrow 2$ must divide $r$ and $r$ is divisible by 6

Possible values of $r$ so that integral terms are obtained i.e., 0, 6, 12, ... 822

Now, this is an A.P. so let $n$ be number of integral terms

$∴$ $822 = 0 + (n - 1) 6 \Rightarrow n = 138$

So, 138 integral terms will be there in the expansion.

Q 34 :
In the expansion of $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}, x \neq 0,$ the sum of the coefficients of $x^{3}$ and $x^{- 13}$ is equal to _______ . [2024]

0%

(118)

The given expansion $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}$

$\Rightarrow {(1 + x)}^{2} (1 - x) \frac{{(x^{3} + 3 x^{2} + 3 x + 1)}^{5}}{x^{15}}$

$\Rightarrow \frac{{(1 + x)}^{2} (1 - x) [{(x + 1)}^{3}]^{5}}{x^{15}} \Rightarrow \frac{(1 - x) {(1 + x)}^{17}}{x^{15}}$

$∴$ Coefficient of $x^{3} \Rightarrow x^{18} in (1 - x) {(1 + x)}^{17}$

$\Rightarrow {(1 + x)}^{17} - x {(1 + x)}^{17}$

$\Rightarrow 0 - x ({}^{17}{C_{17}} x^{17}) = {}^{17}{C_{17}} (- 1) = - 1$

and coefficient of $x^{- 13} \Rightarrow x^{2} in (1 - x) {(1 + x)}^{17}$

$\Rightarrow {(1 + x)}^{17} - x {(1 + x)}^{17}$

$\Rightarrow {}^{17}{C_{2}} - {}^{17}{C_{1}} = 17 \times 8 - 17 = 17 \times 7 = 119$

The sum of the coefficient $x^{3}$ and $x^{- 13} = - 1 + 119 = 118$

Q 35 :
The term independent of x in the expansion of ${(\frac{(x + 1)}{(x^{2 / 3} + 1 - x^{1 / 3})} - \frac{(x - 1)}{(x - x^{1 / 2})})}^{10}, x > 1$ , is: [2025]

120
240
210
150

1

2

3

4

(3)

We have, ${(\frac{(x + 1)}{(x^{\frac{2}{3}} + 1 - x^{\frac{1}{3}})} - \frac{(x - 1)}{(x - x^{\frac{1}{2}})})}^{10}$

$= ((x^{\frac{1}{3}} + 1) - {(\frac{\sqrt{x} + 1}{\sqrt{x}}))}^{10} = {(x^{\frac{1}{3}} - \frac{1}{\sqrt{x}})}^{10}$

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

Since, the term is independent of x, then $\frac{10 - r}{3} - \frac{r}{2} = 0$

$\Rightarrow$ (20 – 2r) – 3r = 0 $\Rightarrow$ r = 4.

Hence, the required term is ${}^{10}C_{4} {(- 1)}^{4} = 210$ .

Q 36 :
In the expansion of ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}, n \in N$ , if the ratio of $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$ , then the value of ${}^{n}C_{3}$ is [2025]

4960
4060
1040
2300

1

2

3

4

(4)

Given, ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}, n \in N$

Now, $T_{r + 1} = {}^{n}C_{r} \cdot {(\sqrt[3]{2})}^{n - r} {(\frac{1}{\sqrt[3]{3}})}^{r}$

$∴ T_{15} = {}^{n}C_{14} \cdot {(\sqrt[3]{2})}^{n - 14} {(\frac{1}{\sqrt[3]{3}})}^{14}$

$15^{th}$ term from end = $T_{n - 13}$ from beginning

$\Rightarrow T_{n - 13} = {}^{n}C_{n - 14} \cdot {(\sqrt[3]{2})}^{14} {(\frac{1}{\sqrt[3]{3}})}^{n - 14}$

But $\frac{T_{15}}{T_{n - 13}} = \frac{{}^{n}C_{14} {(\sqrt[3]{2})}^{n - 14} {(\frac{1}{\sqrt[3]{3}})}^{14}}{{}^{n}C_{n - 14} {(\sqrt[3]{2})}^{14} {(\frac{1}{\sqrt[3]{3}})}^{n - 14}} = \frac{1}{6}$

$\Rightarrow {(\sqrt[3]{2})}^{n - 28} \times {(\sqrt[3]{3})}^{n - 28} = 6^{- 1} \Rightarrow {(\sqrt[3]{6})}^{n - 28} = 6^{- 1}$

$\Rightarrow \frac{n - 28}{3} = - 1 \Rightarrow n = 25$

So, ${}^{n}C_{3} = {}^{25}C_{3} = 2300$ .

Q 37 :
The number of integral terms in the expansion of ${(5^{\frac{1}{2}} + 7^{\frac{1}{8}})}^{1016}$ is [2025]

129
127
130
128

1

2

3

4

(4)

$T_{r + 1} = {}^{1016}C_{r} 5^{\frac{1016 - r}{2}} 7^{r / 8}$ , represents the ${(r + 1)}^{th}$ term of ${(5^{1 / 2} + 7^{1 / 8})}^{1016}$ .

As $T_{r + 1}$ represent an integral term when r is a multiple of 8

i.e., r = 0, 8, 16, 24, ....., 1016

Now, 1016 = 0 + (n – 1)8 [As this form an A.P.]

$\Rightarrow n - 1 = 127 \Rightarrow n = 128$

So, 128 terms are integral terms.

Q 38 :
For some $n \neq 10$ , let the coefficients of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the binomial expansion of ${(1 + x)}^{n + 4}$ be in A.P. Then the largest coefficient in the expansion of ${(1 + x)}^{n + 4}$ is: [2025]

35
20
10
70

1

2

3

4

(1)

Coefficient of $5^{th}, 6^{th}$ and $7^{th}$ terms of the expansion of ${(1 + x)}^{n + 4}$ are ${}^{n + 4}C_{4}, {}^{n + 4}C_{5}$ and ${}^{n + 4}C_{6}$ respectively. As ${}^{n + 4}C_{4}, {}^{n + 4}C_{5}$ and ${}^{n + 4}C_{6}$ are in A.P.

$\Rightarrow 2 \times {}^{n + 4}C_{5} = {}^{n + 4}C_{4} + {}^{n + 4}C_{6}$

$\Rightarrow \frac{2}{5 (n - 1)} = \frac{1}{n (n - 1)} + \frac{1}{30} \Rightarrow \frac{2}{5 (n - 1)} = \frac{30 + n^{2} - n}{30 n (n - 1)}$

$\Rightarrow 12 n = 30 + n^{2} - n \Rightarrow n^{2} - 13 n + 30 = 0$

$\Rightarrow (n - 3) (n - 10) = 0 \Rightarrow n = 3$ [ $∵ n \neq 10$ ]

Here, n + 4 = 3 + 4 = 7

$∴$ Largest binomial coefficient in expansion ${(1 + x)}^{7}$ = Coefficient of middle term = ${}^{7}C_{4} or {}^{7}C_{3} = 35$ .

Q 39 :
The least value of n for which the number of integral terms in the Binomial expansion of ${(\sqrt[3]{7} + \sqrt[12]{11})}^{n}$ is 183, is : [2025]

2148
2184
2196
2172

1

2

3

4

(2)

General term = ${}^{n}C_{r} {(7^{1 / 3})}^{n - r} {(11^{1 / 12})}^{r}$

= ${}^{n}C_{r} {(7)}^{\frac{n - r}{3}} {(11)}^{r / 12}$

For integral terms, r must be multiple of 12

$∴$ r = 12k, k = 0, 1, 2, .....

Total number of values of r = 183

Hence, max r = 12(182) = 2184

Minimum value of n = 2184.

Topic Question Set

Q 31 :
If the second, third and fourth terms in the expansion of ${(x + y)}^{n}$ are 135, 30, and $\frac{10}{3},$ respectively, then $6 (n^{3} + x^{2} + y)$ is equal to ______. [2024]

0%

Q 32 :
The coefficient of $x^{2012}$ in the expansion of ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$ is equal to _______ . [2024]

0%

Q 33 :
Number of integral terms in the expansion of ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$ is equal to _______ . [2024]

0%

Q 34 :
In the expansion of $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}, x \neq 0,$ the sum of the coefficients of $x^{3}$ and $x^{- 13}$ is equal to _______ . [2024]

0%

Q 35 :
The term independent of x in the expansion of ${(\frac{(x + 1)}{(x^{2 / 3} + 1 - x^{1 / 3})} - \frac{(x - 1)}{(x - x^{1 / 2})})}^{10}, x > 1$ , is: [2025]

Q 36 :
In the expansion of ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}, n \in N$ , if the ratio of $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$ , then the value of ${}^{n}C_{3}$ is [2025]

Q 37 :
The number of integral terms in the expansion of ${(5^{\frac{1}{2}} + 7^{\frac{1}{8}})}^{1016}$ is [2025]

Q 38 :
For some $n \neq 10$ , let the coefficients of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the binomial expansion of ${(1 + x)}^{n + 4}$ be in A.P. Then the largest coefficient in the expansion of ${(1 + x)}^{n + 4}$ is: [2025]

Q 39 :
The least value of n for which the number of integral terms in the Binomial expansion of ${(\sqrt[3]{7} + \sqrt[12]{11})}^{n}$ is 183, is : [2025]

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Topic Question Set

Q 31 : If the second, third and fourth terms in the expansion of (x+y)n are 135, 30, and 103, respectively, then 6(n3+x2+y) is equal to ______. [2024]

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Q 32 : The coefficient of x2012 in the expansion of (1-x)2008(1+x+x2)2007 is equal to _______ . [2024]

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Q 33 : Number of integral terms in the expansion of {7(12)+11(16)}824 is equal to _______ . [2024]

0%

Q 34 : In the expansion of (1+x)(1-x2)(1+3x+3x2+1x3)5,x≠0, the sum of the coefficients of x3 and x-13 is equal to _______ . [2024]

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Q 35 : The term independent of x in the expansion of ((x+1)(x2/3+1–x1/3)–(x–1)(x–x1/2))10, x>1, is: [2025]

Q 36 : In the expansion of (23+133)n, n∈N, if the ratio of 15th term from the beginning to the 15th term from the end is 16, then the value of C3n is [2025]

Q 37 : The number of integral terms in the expansion of (512+718)1016 is [2025]

Q 38 : For some n≠10, let the coefficients of the 5th, 6th and 7th terms in the binomial expansion of (1+x)n+4 be in A.P. Then the largest coefficient in the expansion of (1+x)n+4 is: [2025]

Q 39 : The least value of n for which the number of integral terms in the Binomial expansion of (73+1112)n is 183, is : [2025]

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Q 31 :
If the second, third and fourth terms in the expansion of ${(x + y)}^{n}$ are 135, 30, and $\frac{10}{3},$ respectively, then $6 (n^{3} + x^{2} + y)$ is equal to ______. [2024]

Q 32 :
The coefficient of $x^{2012}$ in the expansion of ${(1 - x)}^{2008} {(1 + x + x^{2})}^{2007}$ is equal to _______ . [2024]

Q 33 :
Number of integral terms in the expansion of ${7^{(\frac{1}{2})} + 11^{(\frac{1}{6})}}^{824}$ is equal to _______ . [2024]

Q 34 :
In the expansion of $(1 + x) (1 - x^{2}) {(1 + \frac{3}{x} + \frac{3}{x^{2}} + \frac{1}{x^{3}})}^{5}, x \neq 0,$ the sum of the coefficients of $x^{3}$ and $x^{- 13}$ is equal to _______ . [2024]

Q 35 :
The term independent of x in the expansion of ${(\frac{(x + 1)}{(x^{2 / 3} + 1 - x^{1 / 3})} - \frac{(x - 1)}{(x - x^{1 / 2})})}^{10}, x > 1$ , is: [2025]

Q 36 :
In the expansion of ${(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})}^{n}, n \in N$ , if the ratio of $15^{th}$ term from the beginning to the $15^{th}$ term from the end is $\frac{1}{6}$ , then the value of ${}^{n}C_{3}$ is [2025]

Q 37 :
The number of integral terms in the expansion of ${(5^{\frac{1}{2}} + 7^{\frac{1}{8}})}^{1016}$ is [2025]

Q 38 :
For some $n \neq 10$ , let the coefficients of the $5^{th}, 6^{th}$ and $7^{th}$ terms in the binomial expansion of ${(1 + x)}^{n + 4}$ be in A.P. Then the largest coefficient in the expansion of ${(1 + x)}^{n + 4}$ is: [2025]

Q 39 :
The least value of n for which the number of integral terms in the Binomial expansion of ${(\sqrt[3]{7} + \sqrt[12]{11})}^{n}$ is 183, is : [2025]