The value of $$\frac{100_{C_{50}}}{51}+\frac{100_{C_{51}}}{52}+...+\frac{100_{C_{100}}}{101}$$ is:

We use the identity $$\frac{\binom{n}{r}}{r+1} = \frac{1}{n+1}\binom{n+1}{r+1}$$ which can be verified as follows:

$$\frac{\binom{n}{r}}{r+1} = \frac{n!}{r!(n-r)!} \cdot \frac{1}{r+1} = \frac{n!}{(r+1)!(n-r)!} = \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!(n-r)!} = \frac{1}{n+1}\binom{n+1}{r+1}$$

Applying this identity with $$n = 100$$ shows that each term $$\frac{\binom{100}{r}}{r+1}$$ becomes $$\frac{1}{101}\binom{101}{r+1}$$. Hence, the given sum can be rewritten as

$$S = \sum_{r=50}^{100} \frac{\binom{100}{r}}{r+1} = \frac{1}{101}\sum_{r=50}^{100}\binom{101}{r+1}$$

Setting $$k = r + 1$$ transforms the limits: when $$r = 50$$, $$k = 51$$; when $$r = 100$$, $$k = 101$$. Therefore,

$$S = \frac{1}{101}\sum_{k=51}^{101}\binom{101}{k}$$

Since the binomial coefficients satisfy $$\sum_{k=0}^{101}\binom{101}{k} = 2^{101}$$ and, by symmetry $$\binom{101}{k} = \binom{101}{101-k}$$, the sum over $$k$$ from 51 to 101 equals the sum from 0 to 50. Each half is therefore $$2^{100}$$. It follows that

$$S = \frac{1}{101} \times 2^{100} = \frac{2^{100}}{101}$$

The correct answer is Option A: $$\frac{2^{100}}{101}$$.