Mathematics · 2026
JEE Main · 24 January 2026, Shift 1 · Q7
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
Let $\displaystyle \mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to $\displaystyle 13$ terms. If $\displaystyle 13 \mathrm{~S}=\frac{2^k}{n!}, k \in \mathrm{~N}$, then $\displaystyle n+k$ is equal to
Official answer
From NTA’s final answer key for this paper.
(1)
$\displaystyle 49$
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.