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

Q 21 :
Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

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(5)

$α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$

$= 4 \sum_{r = 1}^{n} r^{2} \frac{n}{r} {}^{n - 1}{C_{r - 1}} + 2 \sum_{r = 1}^{n} r \frac{n}{r} {}^{n - 1}{C_{r - 1}} + \sum_{r = 0}^{n} {}^{n}{C_{r}}$

$= 4 n \sum_{r = 1}^{n} r \cdot {}^{n - 1}{C_{r - 1}} + 2 n \sum_{r = 1}^{n} {}^{n - 1}{C_{r - 1}} + \sum_{r = 0}^{n} {}^{n}{C_{r}}$

$= 4 n (n + 1) 2^{n - 2} + 2 n 2^{n - 1} + 2^{n} = 2^{n} [n (n + 1) + n + 1]$

$= 2^{n} [n^{2} + 2 n + 1] = 2^{n} {(n + 1)}^{2}$

Now, $β = \sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1} + \frac{1}{n + 1} = \sum_{r = 0}^{n} \frac{{}^{n + 1}{C_{r + 1}}}{n + 1} + \frac{1}{n + 1}$

$= \frac{1}{n + 1} \cdot 2^{n + 1} \Rightarrow \frac{2 α}{β} = 2 \cdot \frac{2^{n} {(n + 1)}^{2} \times (n + 1)}{2^{n + 1}} = {(n + 1)}^{3}$

Given that, $140 < \frac{2 α}{β} < 281 \Rightarrow 140 < {(n + 1)}^{3} < 281$

Now, for $n = 4; {(n + 1)}^{3} = 5^{3} = 125$

for $n = 5; {(n + 1)}^{3} = 6^{3} = 216$

for $n = 6; {(n + 1)}^{3} = 7^{3} = 343$

Hence, the value of $n = 5$

Q 22 :
The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

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(1)

$428^{2024} = {(420 + 8)}^{2024}$

$= {}^{2024}{C_{0}} {(420)}^{2024} + {}^{2024}{C_{1}} {(420)}^{2023} {(8)}^{1} + \dots + {}^{2024}{C_{2023}} (420) {(8)}^{2023} + {}^{2024}{C_{2024}} {(8)}^{2024}$

$= 420$ [Some integer] $+ 64^{1012}$

$= 21 \times 20$ [Some integer] $+ {(63 + 1)}^{1012}$

$= 21 \times 20$ [Some integer] $+ 63 \times$ Some integer + 1

$= 21 \times 20$ [Some integer] $+ 21 \times 3 \times$ Some integer + 1

Hence, remainder = 1

Q 23 :
If the coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}; x \neq 0$ is $α,$ then $| α |$ equals ______ . [2024]

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(678)

We have, ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}, x \neq 0$

$= (1 + \frac{6}{x} + \frac{15}{x^{2}} + \frac{20}{x^{3}} + \frac{15}{x^{4}} + \frac{6}{x^{5}} + \frac{1}{x^{6}}) {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}$

$= (1 + \frac{6}{x} + \frac{15}{x^{2}} + \frac{20}{x^{3}} + \frac{15}{x^{4}} + \frac{6}{x^{5}} + \frac{1}{x^{6}}) \times (1 + 7 x^{2} + 21 x^{4} + 35 x^{6} + 35 x^{8} + 21 x^{10} + 7 x^{12} + x^{14})$

$1 - 8 x^{3} + 28 x^{6} - 56 x^{9} + 70 x^{12} - 56 x^{15} + 28 x^{18} - 8 x^{21} + x^{24}$

Coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}$

$= 28 \times 7 + 28 \times 15 - 8 \times 21 \times 6 - 8 \times 20 \times 7 - 8 \times 6 + 15 \times 35 + 35 + 15 \times 21 + 7$

$= 196 + 420 - 1008 - 1120 - 48 + 525 + 35 + 315 + 7$

$= - 678$

$∴$ $| α | = 678$

Q 24 :
If $\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} = \frac{n}{m}$ with $\gcd (n, m) = 1,$ then $n + m$ is equal to _______ . [2024]

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(2041)

$\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10}$

$= ({}^{11}{C_{0}} + \frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} + \frac{{}^{11}{C_{10}}}{11} + \frac{{}^{11}{C_{11}}}{12})$ $- 1 - 1 - \frac{1}{12}$

$= \frac{2^{11 + 1} - 1}{12} - 2 - \frac{1}{12} = \frac{2^{12} - 26}{12} = \frac{4070}{12} = \frac{2035}{6}$

$∴$ $n + m = 2035 + 6 = 2041$

Q 25 :
Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

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(1)

We have, $({{(64)}^{32})}^{32} = ({{(8^{2})}^{32})}^{32} = 8^{2048}$

$= {(9 - 1)}^{2048}$

$= 9^{2048} - {}^{2048}{C_{1}} 9^{2047} + \dots + {}^{2048}{C_{2048} = 9 k + 1}$

So, remainder = 1

Q 26 :
Let $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2}) .$ If $5 α = 6 β,$ then $n$ equals ________ . [2024]

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(10)

We have, $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2})$

Now, $α = \sum_{k = 0}^{n} \frac{{}^{n}{C_{k}}}{k + 1} \cdot {}^{n}{C_{k}}$

$= \sum_{k = 0}^{n} \frac{{}^{n + 1}{C_{k + 1}}}{n + 1} \cdot {}^{n}{C_{n - k}} = \frac{1}{n + 1} \sum_{k = 0}^{n} {}^{n + 1}{C_{k + 1} \cdot {}^{n}{C_{n - k}}}$

$= \frac{1}{n + 1} \cdot {}^{2 n + 1}{C_{n + 1}}$

Now, $β = \sum_{k = 0}^{n - 1} {}^{n}{C_{n - k} \cdot} \frac{n + 1}{(k + 2) (n + 1)} \cdot {}^{n}{C_{k + 1}}$

$= \frac{1}{n + 1} \sum_{k = 0}^{n - 1} {}^{n}{C_{n - k} \cdot {}^{n + 1}{C_{k + 2}}} = \frac{1}{n + 1} \cdot {}^{2 n + 1}{C_{n + 2}}$

$∴$ $\frac{β}{α} = \frac{{}^{2 n + 1}{C_{n + 2}}}{{}^{2 n + 1}{C_{n + 1}}} = \frac{(2 n + 1 - n - 1)! \cdot (n + 1)!}{(2 n + 1 - n - 2)! \cdot (n + 2)!}$

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

$\Rightarrow n = 10$

Q 27 :
If $\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) \cdot {}^{11}C_{r + 1} = \frac{α^{11} - 11^{11}}{10^{10}}$ , then $α$ is equal to : [2025]

11
15
20
24

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(3)

Consider $\frac{10^{r + 1} - 1}{10^{r}} = \frac{10^{r + 1}}{10^{r}} - \frac{1}{10^{r}} = 10 - \frac{1}{10^{r}}$

$∴ \sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) {}^{11}C_{r + 1} = \sum_{r = 0}^{10} (10 - \frac{1}{10^{r}}) {}^{11}C_{r + 1}$

$= \sum_{r = 0}^{10} 10 {}^{11}C_{r + 1} - \sum_{r = 0}^{10} \frac{1}{10^{r}} {}^{11}C_{r + 1}$

Now, $\sum_{r = 0}^{10} 10 {}^{11}C_{r + 1} = 10 \sum_{r = 0}^{10} {}^{11}C_{r + 1} + {}^{11}C_{0} - {}^{11}C_{0}$

$= 10 (2^{11} - 1)$ ... (i)

Also, $\sum_{r = 0}^{10} \frac{1}{10^{r}} {}^{11}C_{r + 1} = \sum_{k = 1}^{11} \frac{1}{10^{k - 1}} {}^{11}C_{k} = 10 \sum_{k = 1}^{11} \frac{1}{10^{k}} {}^{11}C_{k}$ ... (ii)

Consider ${(1 + \frac{1}{10})}^{11} = \sum_{k = 0}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k}$

$= {}^{11}C_{0} + \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k} = 1 + \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k}$

$\Rightarrow \sum_{k = 1}^{11} {}^{11}C_{k} {(\frac{1}{10})}^{k} = {(\frac{11}{10})}^{11} - 1$

Substituting in (ii), we get

$\sum_{r = 0}^{10} {}^{11}C_{r + 1} {(\frac{1}{10})}^{r} = 10 ({(\frac{11}{10})}^{11} - 1)$ ... (iii)

On combining (i) and (iii), we get

$\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) {}^{11}C_{r + 1} = 10 (2^{11} - 1) - 10 ({(\frac{11}{10})}^{11} - 1)$

$= 10 \times 2^{11} - 10 - \frac{11^{11}}{10^{10}} + 10$

$= 10 \times 2^{11} - \frac{11^{11}}{10^{10}}$

$= \frac{{(20)}^{11} - 11^{11}}{10^{10}} \Rightarrow α = 20$ .

Q 28 :
If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

27
18
81
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(3)

We have, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$

Now, $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{r}{2^{r}}) \cdot {}^{9}C_{r} + \sum_{r = 1}^{9} (\frac{3}{2^{r}}) \cdot {}^{9}C_{r}$

$= \sum_{r = 1}^{9} (\frac{9}{2^{r}}) \cdot {}^{8}C_{r - 1} + 3 \sum_{r = 1}^{9} {}^{9}C_{r} {(\frac{1}{2})}^{r}$

$= \frac{9}{2} \sum_{r = 1}^{9} {}^{8}C_{r - 1} {(\frac{1}{2})}^{r - 1} + 3 (\sum_{r = 0}^{9} ({}^{9}C_{r} {(\frac{1}{2})}^{r}) - 1)$

$= \frac{9}{2} {(1 + \frac{1}{2})}^{8} + 3 ({(1 + \frac{1}{2})}^{9} - 1)$

$= \frac{9}{2} \cdot {(\frac{3}{2})}^{8} + 3 {(\frac{3}{2})}^{9} - 3$

$= 6 \cdot {(\frac{3}{2})}^{9} - 3$

On comparing, we get

$\Rightarrow α = 6 and β = 3$

$∴ {(α + β)}^{2} = 81$ .

Q 29 :
The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

3763
18817
16923
33845

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(2)

We have,

${(2 + \sqrt{3})}^{8} = {}^{8}C_{0} 2^{8} + {}^{8}C_{1} 2^{7} (\sqrt{3}) + . . . + {}^{8}C_{8} {(\sqrt{3})}^{8}$

But, for rational terms, we take only those terms whose exponent is an even number.

Required sum = ${}^{8}C_{0} {(2)}^{8} + {}^{8}C_{2} {(2)}^{6} \cdot {(\sqrt{3})}^{2} + {}^{8}C_{4} {(2)}^{4} {(\sqrt{3})}^{4} + {}^{8}C_{6} {(2)}^{2} {(\sqrt{3})}^{6} + {}^{8}C_{8} {(\sqrt{3})}^{8}$

= $2^{8} + 28 \times 2^{6} \times 3 + 70 \times 2^{4} \times 9 + 28 \times 2^{2} \times 27 + 81$

= 256 + 5376 + 10080 + 3024 + 81 = 18817.

Q 30 :
If $1^{2} \cdot ({}^{15}C_{1}) + 2^{2} \cdot ({}^{15}C_{2}) + 3^{2} \cdot ({}^{15}C_{3}) + . . . + 15^{2} \cdot ({}^{15}C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where m, n, k $\in$ N, then m + n + k is equal to : [2025]

21
18
20
19

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3

4

(4)

Using formula, $\sum_{r = 1}^{n} r^{2} {}^{n}C_{r} = n (n + 1) 2^{n - 2}$

For n = 15

$\sum_{r = 1}^{15} r^{2} {}^{15}C_{r} = 15 \times 16 \times 2^{13}$

= $3 \times 5 \times 2^{4} \times 2^{13}$

= $2^{17} \times 3^{1} \times 5^{1}$

On comparing terms, we get m = 17, n = 1 and k = 1

Thus, m + n + k = 19.

Topic Question Set

Q 21 :
Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

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Q 22 :
The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

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Q 23 :
If the coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}; x \neq 0$ is $α,$ then $| α |$ equals ______ . [2024]

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Q 24 :
If $\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} = \frac{n}{m}$ with $\gcd (n, m) = 1,$ then $n + m$ is equal to _______ . [2024]

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Q 25 :
Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

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Q 26 :
Let $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2}) .$ If $5 α = 6 β,$ then $n$ equals ________ . [2024]

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Q 27 :
If $\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) \cdot {}^{11}C_{r + 1} = \frac{α^{11} - 11^{11}}{10^{10}}$ , then $α$ is equal to : [2025]

Q 28 :
If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

Q 29 :
The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

Q 30 :
If $1^{2} \cdot ({}^{15}C_{1}) + 2^{2} \cdot ({}^{15}C_{2}) + 3^{2} \cdot ({}^{15}C_{3}) + . . . + 15^{2} \cdot ({}^{15}C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where m, n, k $\in$ N, then m + n + k is equal to : [2025]

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Topic Question Set

Q 21 : Let α=∑r=0n(4r2+2r+1)Crn and β=(∑r=0nCrnr+1)+1n+1. If 140<2αβ<281, then the value of n is ________. [2024]

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Q 22 : The remainder when 4282024 is divided by 21 is ______ . [2024]

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Q 23 : If the coefficient of x30 in the expansion of (1+1x)6(1+x2)7(1-x3)8; x≠0 is α, then |α| equals ______ . [2024]

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Q 24 : If C1112+C2113+⋯+C91110=nm with gcd(n,m)=1, then n+m is equal to _______ . [2024]

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Q 25 : Remainder when 643232 is divided by 9 is equal to _____ . [2024]

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Q 26 : Let α=∑k=0n((Ckn)2k+1) and β=∑k=0n-1(Ck Ck+1nnk+2). If 5α=6β, then n equals ________ . [2024]

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Q 27 : If ∑r=010(10r+1–110r)·Cr+111=α11–11111010, then α is equal to : [2025]

Q 28 : If ∑r=19(r+32r)·Cr9=α(32)9–β, α,β∈ℕ, then (α+β)2 is equal to [2025]

Q 29 : The sum of all rational terms in the expansion of (2+3)8 is equal to [2025]

Q 30 : If 12·(C115)+22·(C215)+32·(C315)+...+152·(C1515)=2m·3n·5k, where m, n, k ∈ N, then m + n + k is equal to : [2025]

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Q 21 :
Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

Q 22 :
The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

Q 23 :
If the coefficient of $x^{30}$ in the expansion of ${(1 + \frac{1}{x})}^{6} {(1 + x^{2})}^{7} {(1 - x^{3})}^{8}; x \neq 0$ is $α,$ then $| α |$ equals ______ . [2024]

Q 24 :
If $\frac{{}^{11}{C_{1}}}{2} + \frac{{}^{11}{C_{2}}}{3} + \dots + \frac{{}^{11}{C_{9}}}{10} = \frac{n}{m}$ with $\gcd (n, m) = 1,$ then $n + m$ is equal to _______ . [2024]

Q 25 :
Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

Q 26 :
Let $α = \sum_{k = 0}^{n} (\frac{{({}^{n}{C_{k}})}^{2}}{k + 1})$ and $β = \sum_{k = 0}^{n - 1} (\frac{{}^{n}{C_{k} {}^{n}{C_{k + 1}}}}{k + 2}) .$ If $5 α = 6 β,$ then $n$ equals ________ . [2024]

Q 27 :
If $\sum_{r = 0}^{10} (\frac{10^{r + 1} - 1}{10^{r}}) \cdot {}^{11}C_{r + 1} = \frac{α^{11} - 11^{11}}{10^{10}}$ , then $α$ is equal to : [2025]

Q 28 :
If $\sum_{r = 1}^{9} (\frac{r + 3}{2^{r}}) \cdot {}^{9}C_{r} = α {(\frac{3}{2})}^{9} - β, α, β \in ℕ$ , then ${(α + β)}^{2}$ is equal to [2025]

Q 29 :
The sum of all rational terms in the expansion of ${(2 + \sqrt{3})}^{8}$ is equal to [2025]

Q 30 :
If $1^{2} \cdot ({}^{15}C_{1}) + 2^{2} \cdot ({}^{15}C_{2}) + 3^{2} \cdot ({}^{15}C_{3}) + . . . + 15^{2} \cdot ({}^{15}C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where m, n, k $\in$ N, then m + n + k is equal to : [2025]