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

We have, $$S=\dfrac{1}{25!}+\dfrac{1}{3!23!}+\dfrac{1}{5!21!}+...+\dfrac{1}{23!3!}+\dfrac{1}{25!1!}$$ 

or, $$26!\times\ S=26!\left[\dfrac{1}{25!}+\dfrac{1}{3!23!}+\dfrac{1}{5!21!}+...+\dfrac{1}{23!3!}+\dfrac{1}{25!1!} \right]$$ 

or, $$\left[26!\times\ S=\dfrac{26!}{1!25!}+\dfrac{26!}{3!23!}+\dfrac{26!}{5!21!}+...\ +\dfrac{26!}{25!1!}\right]$$

or, $$26!\times\ S=\displaystyle \binom{26}{1}+\binom{26}{3}+\binom{26}{5}+...+\binom{26}{23}+\binom{26}{25}=2^{25}$$ [Since, $$\binom{n}{1}+\binom{n}{3}+...\ =2^{n-1}$$ ]

So, $$26S=\dfrac{2^{25}}{25!}$$ 

or, $$13S=\dfrac{2^{25}}{2\times25!}=\dfrac{2^{24}}{25!}$$

Comparing this to $$13S=\dfrac{2^k}{n!} $$ we have $$k= 24,$$ and $$n=25$$ .

So, $$k+n=24+25=49$$