The value of $$\frac{1}{1!50!} + \frac{1}{3!48!} + \frac{1}{5!46!} + \ldots + \frac{1}{49!2!} + \frac{1}{51!1!}$$ is

Write the given series in a systematic way. Let the index $$k$$ run through the odd integers.

For $$k=1,3,5,\dots ,49$$ we have the terms
$$\frac{1}{k!\,(51-k)!}$$ because $$k+(51-k)=51$$.
There is one additional term
$$\frac{1}{51!\,1!}=\frac{1}{51!}$$.

Convert every term of the first group with the relation
$$\frac{1}{k!\,(51-k)!}= \frac{{}^{51}C_{k}}{51!}$$ because $${}^{51}C_{k}= \dfrac{51!}{k!\,(51-k)!}$$.

Hence the series can be written as
$$S=\sum_{\substack{k=1\\k\;\text{odd}}}^{49} \frac{{}^{51}C_{k}}{51!}\;+\;\frac{1}{51!}$$ or, combining the common denominator $$51!$$,
$$S=\frac{1}{51!}\left(\sum_{\substack{k=1\\k\;\text{odd}}}^{49} {}^{51}C_{k}\;+\;1\right).$$

Recall the binomial identity: for any integer $$n\ge 1$$,
$$\sum_{k\;\text{odd}} {}^{n}C_{k}=2^{\,n-1}.$$ For $$n=51$$ (which is odd) this gives
$$\sum_{k\;\text{odd}} {}^{51}C_{k}=2^{50}.$$

The above sum includes the term $$k=51$$, so
$$\sum_{\substack{k=1\\k\;\text{odd}}}^{49} {}^{51}C_{k}=2^{50}-{}^{51}C_{51}=2^{50}-1$$ because $${}^{51}C_{51}=1$$.

Substitute this result in $$S$$:
$$S=\frac{1}{51!}\Bigl((2^{50}-1)+1\Bigr)=\frac{2^{50}}{51!}.$$

Therefore the value of the series is $$\displaystyle\frac{2^{50}}{51!}$$, which corresponds to Option B.