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

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

Accepted Answer

$\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}C_{1} + \ ^{26}C_{3} + \ ^{26}C_{5} + \ldots. + 13 \right.\ $ terms $)$

$${\frac{1}{26!}\left( \ ^{26}C_{1} + \ ^{26}C_{5} + \ldots. + \ ^{26}C_{25} \right) }{\Rightarrow S = \frac{1}{26!} \times 2^{25} }{\Rightarrow 13\text{ }S = \frac{2^{24}}{25!} }$$so $n + k = 25 + 24 = 49$

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