If $$\frac{1}{n+1}$$ $$^nC_n + \frac{1}{n}$$ $$^nC_{n-1} + \ldots + \frac{1}{2}$$ $$^nC_1 + ^nC_0 = \frac{1023}{10}$$, then $$n$$ is equal to

Explanation

Consider the given series

$$\binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+\cdots+\frac{1}{n}\binom{n}{n-1}+\frac{1}{n+1}\binom{n}{n}=\frac{1023}{10}$$

Observe that each term has the form

$$\frac{1}{r+1}\binom{n}{r}$$

Use the identity

$$\frac{1}{r+1}\binom{n}{r}=\frac{1}{n+1}\binom{n+1}{r+1}$$

Applying this identity to every term,

$$LHS=\frac{1}{n+1}\left[\binom{n+1}{1}+\binom{n+1}{2}+\cdots+\binom{n+1}{n+1}\right]$$

Now use the binomial identity

$$\sum_{r=0}^{n+1}\binom{n+1}{r}=2^{,n+1}$$

Therefore,

$$\binom{n+1}{1}+\binom{n+1}{2}+\cdots+\binom{n+1}{n+1}=2^{,n+1}-1$$

Hence,

$$LHS=\frac{2^{,n+1}-1}{n+1}$$

Given that

$$\frac{2^{,n+1}-1}{n+1}=\frac{1023}{10}$$

Comparing denominators suggests

$$n+1=10$$

Substituting,

$$2^{10}-1=1024-1=1023$$

which matches the numerator exactly.

Therefore,

$$n+1=10$$

$$n=9$$

Hence, the correct answer is

$$\boxed{9}$$