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

Q 1 :
If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of ${(1 + x)}^{n}$ are in the arithmetic progression, then the maximum value of $n$ is: [2024]

14
21
7
28

1

2

3

4

(A)

As ${(1 + x)}^{n} = {}^{n}C_{0} + {}^{n}{C_{1}} x^{1} + {}^{n}{C_{2}} x^{2} + \dots + {}^{n}{C_{n}} x^{n}$

$∴$ ${}^{n}{C_{5}} - {}^{n}{C_{4}} = {}^{n}{C_{6}} - {}^{n}{C_{5}} [Since, {}^{n}{C_{4}}, {}^{n}{C_{5}} and {}^{n}{C_{6}} are in A.P.]$

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

$\Rightarrow \frac{n - 9}{5! (n - 5)! (n - 4)} = \frac{n - 11}{6! (n - 6)! (n - 5)}$

$\Rightarrow 6 (n - 9) = (n - 11) (n - 4)$

$\Rightarrow n^{2} - 21 n + 98 = 0 \Rightarrow n = \frac{21 \pm \sqrt{441 - 392}}{2} = 14, 7$

$∴$ $n_{max} = 14$

Q 2 :
The coefficient of $x^{70}$ in $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + x^{4} {(1 + x)}^{96} + \dots + x^{54} {(1 + x)}^{46}$ is ${}^{99}{C_{p} - {}^{46}{C_{q} .}}$ Then a possible value of $p + q$ is: [2024]

68
83
55
61

1

2

3

4

(B)

We have, $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + \dots + x^{54} {(1 + x)}^{46}$

It is a G.P. with first term $= x^{2} {(1 + x)}^{98}$

and common ratio $= \frac{x}{1 + x}$

$∴$ Sum of these terms $= x^{2} {(1 + x)}^{98} (\frac{{(\frac{x}{1 + x})}^{53} - 1}{\frac{x}{1 + x} - 1})$

$= x^{2} {(1 + x)}^{98} ((1 + x) - x^{53} {(1 + x)}^{- 52})$

$= x^{2} \underset{coeff . of x^{68}}{\underset{⏝}{{(1 + x)}^{99}}}$ $- x^{55}$ $\underset{coeff . of x^{15}}{\underset{⏝}{{(1 + x)}^{46}}}$ [ $∵$ we need coefficient of $x^{70}$ ]

$= {}^{99}{C_{68}} - {}^{46}{C_{15}} = {}^{99}{C_{p}} - {}^{46}{C_{q}}$ ( $∵$ Given)

Hence, $p + q = 83$

Q 3 :
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]

0%

(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 4 :
The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

0%

(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 5 :
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]

0%

(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 6 :
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]

0%

(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 7 :
Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

0%

(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 8 :
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]

0%

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

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

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

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

$\Rightarrow n = 10$

Topic Question Set

Q 1 :
If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of ${(1 + x)}^{n}$ are in the arithmetic progression, then the maximum value of $n$ is: [2024]

Q 2 :
The coefficient of $x^{70}$ in $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + x^{4} {(1 + x)}^{96} + \dots + x^{54} {(1 + x)}^{46}$ is ${}^{99}{C_{p} - {}^{46}{C_{q} .}}$ Then a possible value of $p + q$ is: [2024]

Q 3 :
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 4 :
The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

0%

Q 5 :
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]

0%

Q 6 :
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]

0%

Q 7 :
Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

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Q 8 :
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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Topic Question Set

Q 1 : If the coefficients of x4,x5 and x6 in the expansion of (1+x)n are in the arithmetic progression, then the maximum value of n is: [2024]

Q 2 : The coefficient of x70 in x2(1+x)98+x3(1+x)97+x4(1+x)96+…+x54(1+x)46 is Cp-Cq.4699 Then a possible value of p+q is: [2024]

Q 3 : 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 4 : The remainder when 4282024 is divided by 21 is ______ . [2024]

0%

Q 5 : 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 6 : If C1112+C2113+⋯+C91110=nm with gcd(n,m)=1, then n+m is equal to _______ . [2024]

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Q 7 : Remainder when 643232 is divided by 9 is equal to _____ . [2024]

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Q 8 : 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 1 :
If the coefficients of $x^{4}, x^{5}$ and $x^{6}$ in the expansion of ${(1 + x)}^{n}$ are in the arithmetic progression, then the maximum value of $n$ is: [2024]

Q 2 :
The coefficient of $x^{70}$ in $x^{2} {(1 + x)}^{98} + x^{3} {(1 + x)}^{97} + x^{4} {(1 + x)}^{96} + \dots + x^{54} {(1 + x)}^{46}$ is ${}^{99}{C_{p} - {}^{46}{C_{q} .}}$ Then a possible value of $p + q$ is: [2024]

Q 3 :
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 4 :
The remainder when $428^{2024}$ is divided by 21 is ______ . [2024]

Q 5 :
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 6 :
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 7 :
Remainder when $64^{32^{32}}$ is divided by 9 is equal to _____ . [2024]

Q 8 :
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]