If C _ r ≡ C _ r 25 and C _ 0 + 5 ∙ C _ 1 + 9 ∙ C _ 2 + + (101) ∙ C _ 25 = 2 ^ 25 ∙ k, then k is equal to ____________.

Correct answer: 0 ∑ _ r = 0 ^ 25 (4r + 1) ^ 25 C _ r = 4 ∑ _ r = 0 ^ 25 r . ^ 25 C _ r + ∑ _ r = 0 ^ 25 Cr 25 = 4 ∑ _ r = 1 ^ 25 r × (25)/(r) Cr - 1 24 + 2 ^ 25 = 100 ∑ _ r = 1 ^ 25 Cr - 1 25 + 2 ^ 25 = 100.2 ^ 24 + 2 ^ 25 = 2 ^ 25 (50 + 1) = 51 .2 ^ 25 So, k = 51

JEE Main 20209 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2020 · 9 January · Evening Shift · Mathematics · Binomial Theorem

If C _ r ≡ C _ r 25 and C _ 0 + 5 ∙ C _ 1 + 9 ∙ C _ 2 + + (101) ∙ C...

{C}_{r} ≡ {C}_{r} 25

and

{C}_{0} + 5 ∙ {C}_{1} + 9 ∙ {C}_{2} + \ldots + \left(101\right) ∙ {C}_{25} = {2}^{25} ∙ k,

then

k

is equal to ____________.

Correct answer

Step-by-step solution

$\mathop{∑}_{r = 0}^{25}{\left(4r + 1\right)}^{25}{C}_{r}$

$= 4\mathop{∑}_{r = 0}^{25}r{.}^{25}{C}_{r} + \mathop{∑}_{r = 0}^{25} Cr 25$
$= 4\mathop{∑}_{r = 1}^{25}r \times \dfrac{25}{r} Cr - 1 24 + {2}^{25}$

$= 100\mathop{∑}_{r = 1}^{25} Cr - 1 25 + {2}^{25}$
$= {100.2}^{24} + {2}^{25}$

$= {2}^{25}\left(50 + 1\right)$

$= 51{.2}^{25}$
So, $k = 51$

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