Let S =1/25!+1/3!23!+1/5!21!+ldots up to 13 terms. If 13 S =frac 2 k n ! , k in N , then n + k is equal to

Mathematics · Binomial Theorem

JEE Main 2026 — 24 January, Morning Shift — Question 14

Let $\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13 \mathrm{~S}=\frac{2^{\mathrm{k}}}{\mathrm{n}!}, \mathrm{k} \in \mathrm{N}$ , then $\mathrm{n}+\mathrm{k}$ is equal to

Option A:
$51$
Option B:
$52$
Option C:
$49$
Correct
Option D:
$50$

Answer: C

Step-by-step solution

$\frac{1}{26!}\left(\frac{26!}{25!1!}+\frac{26!}{3!23!}+\frac{26!}{5!21!}+\ldots . .+13\right.$ terms $)$

$\frac{1}{26!}\left({ }^{26} \mathrm{C}_{1}+{ }^{26} \mathrm{C}_{3}+{ }^{26} \mathrm{C}_{5}+\ldots .+13\right.$ terms $)$

$\frac{1}{26!}\left({ }^{26} \mathrm{C}_{1}+{ }^{26} \mathrm{C}_{5}+\ldots .+{ }^{26} \mathrm{C}_{25}\right)$

$\mathrm{S}=\frac{1}{26!} \times 2^{25}$

$\Rightarrow 13 \mathrm{~S}=\frac{2^{24}}{25!}$

so $n+k=25+24=49$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients