Proof with binomial coefficients and induction I am stuck and unable to prove the following expression with induction
$$ \sum_{k=0}^{n} \binom{2 n}{2 k}= 2^{2n-1} $$
where $n>0$. One has to show that the equality below is satisfied:
$$ \sum_{k=0}^{n+1} \binom{2n+2}{2k} = 2^{2n+1}. $$
I have tried to show with binomial identities and index moving that one can come from $\binom{2n+2}{2k}$ to $\binom{2n}{2k}$.
Unfortunately I cannot move on with this.
Thanks in advance for your help
 A: How $\binom{m}{p}=\binom{m-1}{p-1}+\binom{m-1}{p}$, if $m=2n$ and $p=2k$, then:
$$\binom{2n+2}{2k}=\binom{2n}{2k-2}+2\cdot\binom{2n}{2k-1}+\binom{2n}{2k}$$
Therefore
\begin{eqnarray*}{}\sum_{k=0}^{n+1}\binom{2n+2}{2k}&=&\sum_{k=0}^{n+1}\binom{2n}{2k-2}+2\sum_{k=0}^{n+1}\binom{2n}{2k-1}+\sum_{k=0}^{n+1}\binom{2n}{2k}\\
&=&\sum_{k=1}^{n+1}\binom{2n}{2k-2}+\binom{2n}{-2}+2\sum_{k=1}^{n}\binom{2n}{2k-1}+2\binom{2n}{-1}+2\binom{2n}{2n+1}+\sum_{k=0}^{n}\binom{2n}{2k}+\binom{2n}{2n+2}\end{eqnarray*}
Take $\binom{m}{p}=0$, if $p<0$ or $p>m$, then:
$$\sum_{k=0}^{n+1}\binom{2n+2}{2k}
=\sum_{k=1}^{n+1}\binom{2n}{2k-2}+2\sum_{k=1}^{n}\binom{2n}{2k-1}+\sum_{k=0}^{n}\binom{2n}{2k}$$
How
$$\sum_{k=0}^{n}\binom{2n}{2k}=\sum_{k=1}^{n}\binom{2n}{2k-1}$$
And
$$\sum_{k=0}^{n}\binom{2n}{2k}=\sum_{k=1}^{n+1}\binom{2n}{2k-2}$$
Then
$$\sum_{k=0}^{n+1}\binom{2n+2}{2k}
=\sum_{k=1}^{n+1}\binom{2n}{2k-2}+2\sum_{k=1}^{n}\binom{2n}{2k-1}+\sum_{k=0}^{n}\binom{2n}{2k}=4\sum_{k=0}^{n}\binom{2n}{2k}=2^2\cdot2^{2n-1}=2^{2n+1}$$
