What is wrong with the proof? I try to prove the identity:
$$\sum_{k=0}^n \binom{2n+1}{2k}=2^{2n}$$
but have troubles with the last step:
$$\sum_{k=0}^n \binom{2n+1}{k}=\sum_{k=0}^n  \left( \binom{2n}{2k} + \binom{2n}{2k-1}\right) =\sum_{j=-1}^{2n}\binom{2n}{j}=2^{2n}$$
But should not $j$ start from $0$? Is something wrong?
 A: Following the comment of Nicolas and the partial develop of Tyler:
$$\sum_{k=0}^n \binom{2n+1}{2k}=\binom{2n+1}{0}+\sum_{k=1}^n  \left( \binom{2n}{2k} + \binom{2n}{2k-1}\right) =1+\color{red}{\sum_{k=1}^{n}\binom{2n}{2k}}+\sum_{k=1}^n\binom{2n}{2k-1}$$
Observe that the marked term on red are the even coefficients, but $2k=0$, of a complete binomial sum of the kind
$$\sum_{k=0}^{2n}\binom{2n}{k}=2^{2n}=\sum_{k=0}^{n}\binom{2n}{2k}+\sum_{k=1}^{n}\binom{2n}{2k-1}=1+\sum_{k=1}^{n}\binom{2n}{2k}+\sum_{k=1}^n\binom{2n}{2k-1}$$
In other words, what remains to prove are the identities
$$\sum_{k=0\\k\text{ even}}^{2n}\binom{2n}{k}=\sum_{k=0}^{n}\binom{2n}{2k}$$
$$\sum_{k=0\\k\text{ odd}}^{2n}\binom{2n}{k}=\sum_{k=1}^{n}\binom{2n}{2k-1}$$
A: The important identity is
$$
\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}\tag{*}
$$
which makes sense only when $k\ge0$, unless we extend the definition of the binomial coefficient. For arbitrary $n$ and integer $k$, we can set
$$
\binom{n}{k}=
\begin{cases}
\dfrac{n(n-1)\dotsm(n-k+1)}{k!} & \text{if $k\ge0$} \\[6px]
0 & \text{if $k<0$}
\end{cases}
$$
In this way, the identity (*) continues to hold for any $n$ and any integer $k$. Thus we have
$$
\sum_{k=0}^n\binom{2n+1}{2k}=
\sum_{k=0}^n\left(\binom{2n}{2k}+\binom{2n}{2k-1}\right)=
\sum_{\substack{-1\le k\le 2n\\k\text{ even}}}\binom{2n}{k}+
\sum_{\substack{-1\le k\le 2n\\k\text{ odd}}}\binom{2n}{k}\\=
\sum_{-1\le k\le n}\binom{2n}{k}=
\sum_{0\le k\le n}\binom{2n}{k}=2^{2n}
$$
