MCQMediumJEE Main 2026 · 24 January, Shift 1Sum of Series

Mathematics Question from JEE Main 2026 · 24 January, Shift 1

Let S=125!+13!23!+15!21!+S=\frac{1}{25!}+\frac{1}{3!\,23!}+\frac{1}{5!\,21!}+\cdots up to 1313 terms. If 13S=2kn!13S=\dfrac{2^k}{n!}, kNk\in\mathbb{N}, then n+kn+k is equal to

  • A

    5252

  • B

    5151

  • C

    4949

  • D

    5050

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