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

Q 11 :
The largest natural number n such that $3^{n}$ divides 66! is _______ . [2023]

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(31)

The largest prime $p$ in $n!$ is

$E_{p} (n!) = [\frac{n}{p}] + [\frac{n}{p^{2}}] + [\frac{n}{p^{3}}] + \dots$

Here, $p$ = 3 is a prime number and $n$ = 66

$∴$ $E_{3} (66!) = [\frac{66}{3}] + [\frac{66}{9}] + [\frac{66}{27}] + [\frac{66}{81}] + \dots$

$= 22 + 7 + 2 + 0 + \dots = 31$ $∴$ $66! = 3^{31} \dots$

$∴ The maximum exponent of 3 is 31.$

Q 12 :
The remainder, when $7^{103}$ is divided by 17, is ________ . [2023]

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(12)

$7^{103} = 7 . 7^{102}$

$7 . 7^{102} = 7 {(7^{2})}^{51} = 7 {(49)}^{51} = 7 {(51 - 2)}^{51}$

$= 7 {(- 2)}^{51} = - 7 {(2)}^{51} = - 7 (2^{48 + 3}) = - 7.8 (2^{48})$

$= - 56 {(2^{4})}^{12} = - 56 {(17 - 1)}^{12} = - 56 \times {(- 1)}^{12} = (- 56)$

$Remainder = 12$

Q 13 :
The remainder, when $19^{200} + 23^{200}$ is divided by 49, is ________ . [2023]

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(29)

$23^{200} + 19^{200} = {(21 + 2)}^{200} + {(21 - 2)}^{200}$

$= 2 [{}^{200}{C_{0}} 21^{200} 2^{0} + {}^{200}{C_{2}} 21^{198} 2^{2} + \dots + {}^{200}{C_{200}} 21^{0} 2^{200}]$ ...(i)

Now, $21^{2}$ will be divisible by 49, so (i) can be written as $2 [49 λ + 2^{200}]$

$\frac{2 [49 λ + 2^{200}]}{49}$ $then, Remainder = \frac{2^{201}}{49} = \frac{{(8)}^{67}}{49} = \frac{{(7 + 1)}^{67}}{49}$

$= \frac{{}^{67}{C_{0}} 7^{67} \cdot 1^{0} + {}^{67}{C_{1}} 7^{66} \cdot 1 + \dots + {}^{67}{C_{66}} 7 \cdot 1^{66} + {}^{67}{C_{67} \cdot 1}}{49}$

$Remainder = \frac{67 \times 7 + 1}{49}$ $[other terms will be divisible by 49]$

$= \frac{(49 + 18) \times 7 + 1}{49}, Remainder = 29$

Q 14 :
Suppose $\sum_{r = 0}^{2023} r^{2} {}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ___________ . [2023]

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(1012)

$\sum_{r = 0}^{2023} r^{2}$ ${}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$

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

Then, $\sum_{r = 0}^{2023} r^{2} \cdot {}^{2023}{C_{r}} = (2023) (2023 + 1) \cdot 2^{2023 - 2}$

$= 2023 \times 2024 \times 2^{2021} = 2023 \times α \times 2^{2022}$

$∴$ $2 α = 2024 \Rightarrow α = 1012$

Q 15 :
Let the sum of the coefficients of the first three terms in the expansion of ${(x - \frac{3}{x^{2}})}^{n}, x \neq 0, n \in N,$ be 376. Then the coefficient of $x^{4}$ is ___________ . [2023]

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(405)

Sum of coefficients of first three terms of ${(x - \frac{3}{x^{2}})}^{n}$ is 376.

${}^{n}C_{0} - {}^{n}C_{1} \cdot 3 + {}^{n}C_{2} \cdot 3^{2} = 376 \Rightarrow 1 - 3 n + \frac{n (n - 1)}{2} \times 9 = 376$

$\Rightarrow 3 n^{2} - 5 n - 250 = 0 ∴ n = 10$

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

For coefficient of $x^{4}$ $\Rightarrow 10 - 3 r = 4 \Rightarrow r = 2$

Coefficient of $x^{4}$ $= {}^{10}C_{2} {(- 3)}^{2} = 405$

Q 16 :
The remainder when ${(2023)}^{2023}$ is divided by 35 is __________ . [2023]

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(7)

${(2023)}^{2023} = {(2030 - 7)}^{2023} = {(35 K - 7)}^{2023}$

$= {}^{2023}{C_{0}} {(35 K)}^{2023} {(- 7)}^{0} + {}^{2023}{C_{1}} {(35 K)}^{2022} {(- 7)}^{1} + \dots + {}^{2023}{C_{2023}} {(- 7)}^{2023}$

$= 35 N - 7^{2023}$

Now, $- 7^{2023} = - 7 \times 7^{2022} = - 7 {(7^{2})}^{1011} = - 7 {(50 - 1)}^{1011}$

$= - 7 [{}^{1011}{C_{0} (50)}^{1011} - {}^{1011}{C_{1}} {(50)}^{1010} + \dots + {}^{1011}{C_{1011}}]$

$= - 7 (5 λ - 1) = 35 λ + 7$

$∴ When {(2023)}^{2023} is divided by 35, remainder is 7 .$

Q 17 :
50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

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(23)

Given, $x^{1 / 50} = 12 \Rightarrow x = {(12)}^{50}$

$y^{1 / 50} = 18 \Rightarrow y = {(18)}^{50}$

$∴$ $\frac{x + y}{25} = remainder \Rightarrow \frac{x + y}{25} = \frac{12^{50} + 18^{50}}{25}$

$\frac{12^{50}}{25} + \frac{{(18)}^{50}}{25} = \frac{{(144)}^{25} + {(324)}^{25}}{25} = \frac{{(150 - 6)}^{25} + {(325 - 1)}^{25}}{25}$

$= \frac{25 k - (6^{25} + 1)}{25} = \frac{25 k - [{(5 + 1)}^{25} + 1]}{25} = \frac{25 k_{1} - 2}{25}$

$∴$ $Remainder = 23$

Q 18 :
The remainder on dividing $5^{99}$ by 11 is _________ . [2023]

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(9)

$We have, 5^{99} = 5^{4} \times 5^{95} = 625 \times {(5^{5})}^{19}$

$= 625 \times {(3125)}^{19} = 625 \times {(3124 + 1)}^{19}$

$= 625 \times {(11 λ + 1)}^{19} [∵ 11 \times 284 = 3124]$

$∴$ $Remainder = 9$

Q 19 :
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

(1)

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 20 :
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

(2)

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, $p + q = 83$

Topic Question Set

Q 11 :
The largest natural number n such that $3^{n}$ divides 66! is _______ . [2023]

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Q 12 :
The remainder, when $7^{103}$ is divided by 17, is ________ . [2023]

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Q 13 :
The remainder, when $19^{200} + 23^{200}$ is divided by 49, is ________ . [2023]

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Q 14 :
Suppose $\sum_{r = 0}^{2023} r^{2} {}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ___________ . [2023]

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Q 15 :
Let the sum of the coefficients of the first three terms in the expansion of ${(x - \frac{3}{x^{2}})}^{n}, x \neq 0, n \in N,$ be 376. Then the coefficient of $x^{4}$ is ___________ . [2023]

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Q 16 :
The remainder when ${(2023)}^{2023}$ is divided by 35 is __________ . [2023]

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Q 17 :
50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

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Q 18 :
The remainder on dividing $5^{99}$ by 11 is _________ . [2023]

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Q 19 :
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 20 :
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]

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Topic Question Set

Q 11 : The largest natural number n such that 3n divides 66! is _______ . [2023]

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Q 12 : The remainder, when 7103 is divided by 17, is ________ . [2023]

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Q 13 : The remainder, when 19200+23200 is divided by 49, is ________ . [2023]

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Q 14 : Suppose ∑r=02023 r2 Cr2023=2023×α×22022. Then the value of α is ___________ . [2023]

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Q 15 : Let the sum of the coefficients of the first three terms in the expansion of (x-3x2)n, x≠0,n∈N, be 376. Then the coefficient of x4 is ___________ . [2023]

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Q 16 : The remainder when (2023)2023 is divided by 35 is __________ . [2023]

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Q 17 : 50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

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Q 18 : The remainder on dividing 599 by 11 is _________ . [2023]

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Q 19 : 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 20 : 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]

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Q 11 :
The largest natural number n such that $3^{n}$ divides 66! is _______ . [2023]

Q 12 :
The remainder, when $7^{103}$ is divided by 17, is ________ . [2023]

Q 13 :
The remainder, when $19^{200} + 23^{200}$ is divided by 49, is ________ . [2023]

Q 14 :
Suppose $\sum_{r = 0}^{2023} r^{2} {}^{2023}{C_{r}} = 2023 \times α \times 2^{2022}$ . Then the value of $α$ is ___________ . [2023]

Q 15 :
Let the sum of the coefficients of the first three terms in the expansion of ${(x - \frac{3}{x^{2}})}^{n}, x \neq 0, n \in N,$ be 376. Then the coefficient of $x^{4}$ is ___________ . [2023]

Q 16 :
The remainder when ${(2023)}^{2023}$ is divided by 35 is __________ . [2023]

Q 17 :
50th root of a number x is 12 and 50th root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is _________ . [2023]

Q 18 :
The remainder on dividing $5^{99}$ by 11 is _________ . [2023]

Q 19 :
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 20 :
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]