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

We need to find the value of $$S = \frac{1}{1!\cdot 50!} + \frac{1}{3!\cdot 48!} + \frac{1}{5!\cdot 46!} + \ldots + \frac{1}{49!\cdot 2!} + \frac{1}{51!\cdot 1!}$$.

Multiplying and dividing by $$51!$$ yields $$S = \frac{1}{51!}\left[\frac{51!}{1!\cdot 50!} + \frac{51!}{3!\cdot 48!} + \frac{51!}{5!\cdot 46!} + \ldots + \frac{51!}{49!\cdot 2!} + \frac{51!}{51!\cdot 1!}\right].$$

Recognizing that for the terms from $$\frac{51!}{1!\cdot 50!}$$ to $$\frac{51!}{49!\cdot 2!}$$ we have $$\frac{51!}{k!(51-k)!} = \binom{51}{k}$$ where $$k = 1, 3, 5, \ldots, 49$$, and noting that $$\frac{51!}{51!\cdot 1!} = 1$$ and $$\binom{51}{51} = \frac{51!}{51!\cdot 0!} = 1$$, it follows that $$S = \frac{1}{51!}\left[\binom{51}{1} + \binom{51}{3} + \binom{51}{5} + \ldots + \binom{51}{49} + \binom{51}{51}\right].$$

Using the binomial sum identity $$\binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \ldots = 2^{n-1},$$ for $$n = 51$$ we obtain $$\binom{51}{1} + \binom{51}{3} + \binom{51}{5} + \ldots + \binom{51}{51} = 2^{50}.$$

Therefore, $$S = \frac{2^{50}}{51!}.$$

The answer is Option B: $$\frac{2^{50}}{51!}$$.