1. Write the series in a general form

The given sum is

$$S=\frac{1}{1!50!}+\frac{1}{3!48!}+\frac{1}{5!46!}+\cdots+\frac{1}{49!2!}+\frac{1}{51!1!}.$$

Observe that the factorials in each term add up to $51$:

$$1+50=3+48=5+46=\cdots=51+0.$$

But here the listed last terms are $\frac{1}{49!2!}$ and $\frac{1}{51!1!}$, so the intended pattern is:

$$S=\sum_{\substack{r=1 \\ r\text{ odd}}}^{51} \frac{1}{r!(51-r)!}.$$

That is, we are summing over all odd $r$ from $1$ to $51$.

2. Relate to binomial coefficients

Recall:

$$\binom{51}{r}=\frac{51!}{r!(51-r)!}.$$

So,

$$\frac{1}{r!(51-r)!}=\frac{1}{51!}\binom{51}{r}.$$

Hence,

$$S=\frac{1}{51!}\sum_{\substack{r=1 \\ r\text{ odd}}}^{51}\binom{51}{r}.$$

3. Use the standard identity for odd binomial coefficients

For any positive integer $n$,

$$\sum_{r\text{ odd}} \binom{n}{r}=2^{n-1}.$$

Thus for $n=51$,

$$\sum_{r\text{ odd}} \binom{51}{r}=2^{50}.$$

Therefore,

$$S=\frac{1}{51!}\cdot 2^{50}=\frac{2^{50}}{51!}.$$

4. Match with the options

So the value is

$$\boxed{\frac{2^{50}}{51!}}.$$

This is Option D.

5. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

They match.