<s>[INST] Proving $\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{2^n}{n+1}$ using the extraction/absorbtion identity. I want to use the extraction/absorbtion identity to prove the equality
$$
\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{2^n}{n+1}
$$
Followed from this identity, we could directly obtain
$$
\binom{n+1}{2k+1}=\frac{n+1}{2k+1}\binom{n}{2k}
$$
Then we have
$$
\sum_k\binom{n}{2k}\frac{1}{2k+1}=\frac{1}{n+1}\sum_k\binom{n+1}{2k+1}
$$
Then how can I prove that $\sum_k\binom{n+1}{2k+1}=2^n$? Thanks!
 [/INST] Since $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$, we have
$$\binom{n+1}{1} + \binom{n+1}{3} + \binom{n+1}{5} + \dots = \left[\binom{n}{0} + \binom{n}{1}\right] + \left[\binom{n}{2} + \binom{n}{3}\right] + \cdots$$
which is $2^n$.
</s>