Let S=(1)/(25!)+(1)/(3!23!)+(1)/(5!21!)+ up to 13 terms. If 13 ~S= 2^ k n! , k ∈ ~N , then n+k is equal to

Correct answer: B. 49 The general term is (1)/((2r-1)!(27-2r)!) for r=1 to 13 . Multiply and divide by 26! : (1)/((2r-1)!(27-2r)!) = (1)/(26!) 26 2r-1 S = (1)/(26!)∑_ r=1 ^ 13 26 2r-1 = (1)/(26!)[ 26 1 + 26 3 + + 26 25 ] Sum of odd-indexed binomial coefficients: ∑ 26 odd = 2^ 25 S = 2^ 25 26! 13S = 13 · 2^ 25 26! = 13 · 2^ 25 26 · 25! = 2^ 25 2 · 25! = 2^ 24 25! So k = 24,\; n = 25 , giving n + k = 49 .

JEE Main 202624 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2026 · 24 January · Morning Shift · Mathematics · Binomial Theorem

Let S=(1)/(25!)+(1)/(3!23!)+(1)/(5!21!)+ up to 13 terms. If 13 ~S=...

Let

\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots

up to 13 terms. If

13 \mathrm{~S}=\frac{2^{k}}{n!}, k \in \mathrm{~N}

, then

n+k

is equal to

Options

Correct answer

B. 49

Step-by-step solution

The general term is

\frac{1}{(2r-1)!(27-2r)!}

for

r=1

13

.
Multiply and divide by

26!

\frac{1}{(2r-1)!(27-2r)!} = \frac{1}{26!}\binom{26}{2r-1}

S = \frac{1}{26!}\sum_{r=1}^{13}\binom{26}{2r-1} = \frac{1}{26!}\left[\binom{26}{1}+\binom{26}{3}+\cdots+\binom{26}{25}\right]

Sum of odd-indexed binomial coefficients:

\sum \binom{26}{\text{odd}} = 2^{25}

S = \frac{2^{25}}{26!}

13S = \frac{13 \cdot 2^{25}}{26!} = \frac{13 \cdot 2^{25}}{26 \cdot 25!} = \frac{2^{25}}{2 \cdot 25!} = \frac{2^{24}}{25!}

k = 24,\; n = 25

, giving

n + k = 49

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