Prove this sum $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$ Show that
$$\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$$
 A: Hint Let
$$a_{n}=\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\cdot \binom{2n-k+1}{k}=2^{2n}+\sum_{k=1}^{n}(-1)^k\cdot 2^{2n-2k}\left[\binom{2n-k}{k}+\binom{2n-k}{k-1}\right]$$
other hand we easy to prove $r=k-1$
\begin{align*}&\sum_{k=1}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k}{k-1}=\sum_{r=0}^{n-1}(-1)^{r+1}\cdot 2^{2(n-1)-2r}\binom{2(n-1)-r+1}{r}\\&=
-a_{n-1}
\end{align*}
if we let $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k}{k}=b_{n}$
then we have
$$b_{n}=a_{n}+a_{n-1}$$
and
\begin{align*}b_{n}&=2^{2n}+\sum_{k=1}^{n-1}(-1)^k\cdot 2^{2n-2k}\binom{2n-k}{k}+(-1)^n\\
&=2^{2n}+\sum_{k=1}^{n-1}(-1)^k\cdot 2^{2n-2k}\left[\binom{2n-k-1}{k}+\binom{2n-k-1}{k-1}\right]+(-1)^n\\
&=4a_{n-1}-\sum_{j=0}^{n-1}(-1)^j\cdot 2^{2(n-1)-2j}\binom{2(n-1)-j}{j}\\
&=4a_{n-1}-b_{n-1}
\end{align*}
so
$$a_{n}+a_{n-1}=4a_{n-1}-a_{n-1}-a_{n-2}$$
then we have
$$a_{n}+a_{n-2}=2a_{n-1},a_{0}=1,a_{1}=2$$
so
$$a_{n}=n+1$$
A: We can compute the generating function:
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k2^{2n-2k}\binom{2n-k+1}{k}x^{2n+1}
&=\frac12\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k2^{2n-2k+1}\binom{2n-k+1}{2n-2k+1}x^{2n+1}\tag{1}\\
&=\frac12\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{k+1}2^{2n-2k+1}\binom{-k-1}{2n-2k+1}x^{2n+1}\tag{2}\\
&=\frac12\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^{k+1}2^{2n-2k+1}\binom{-k-1}{2n-2k+1}x^{2n+1}\tag{3}\\
&=\frac12\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^{k+1}2^{2n+1}\binom{-k-1}{2n+1}x^{2n+2k+1}\tag{4}\\
&=\frac14\sum_{k=0}^\infty(-1)^k\left(\frac{x^{2k}}{(1-2x)^{k+1}}-\frac{x^{2k}}{(1+2x)^{k+1}}\right)\tag{5}\\
&=\frac14\left(\frac1{1-2x+x^2}-\frac1{1+2x+x^2}\right)\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: $\binom{-n}{k}=(-1)^k\binom{k+n-1}{k}$
$(3)$: switch the order of summation
$(4)$: substitute $n\mapsto n+k$
$(5)$: the sum in $n$ is the odd part of $(-1)^{k+1}\frac{x^{2k}}{(1+2x)^{k+1}}$
$(6)$: sum the geometric series

$(6)$ is the odd part of $\frac12\frac1{(1-x)^2}=\frac12\sum\limits_{n=0}^\infty(n+1)x^n$; that is,
$$
\frac12\sum_{n=0}^\infty((2n+1)+1)x^{2n+1}=\sum_{n=0}^\infty(n+1)x^{2n+1}\tag{7}
$$
Equating coefficients of $x^{2n+1}$, we get
$$
\sum_{k=0}^n(-1)^k2^{2n-2k}\binom{2n-k+1}{k}=n+1\tag{8}
$$
A: Suppose we seek to verify that
$$\sum_{k=0}^n (-1)^k 2^{2n-2k} {2n-k+1\choose k}
= n+1.$$
Introduce
$${2n-k+1\choose k} = {2n-k+1\choose 2n-2k+1} 
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n-k+1}}{z^{2n-2k+2}} \; dz.$$
Observe  that this  is zero  when $k\gt  n$ so  we may  extend $k$  to
infinity.
We thus obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n+1}}{z^{2n+2}}
\sum_{k\ge 0} (-1)^k 2^{2n-2k} \frac{z^{2k}}{(1+z)^{k}} \; dz
\\ = \frac{2^{2n}}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n+1}}{z^{2n+2}}
\frac{1}{1+z^2/(1+z)/4}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n+2}}{z^{2n+2}}
\frac{1}{4(1+z)+z^2}
\; dz
\\ = \frac{2^{2n+2}}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2n+2}}{z^{2n+2}}
\frac{1}{(z+2)^2}
\; dz.$$
Evaluate the integral by computing the negative of the residue
at $z=-2$. This requires
$$\left(\frac{(1+z)^{2n+2}}{z^{2n+2}}\right)'
= \frac{(2n+2)(1+z)^{2n+1}}{z^{2n+2}}
- (2n+2)\frac{(1+z)^{2n+2}}{z^{2n+3}}.$$
Setting $z=-2$ we obtain
$$2^{2n+2} (2n+2) \frac{(-1)^{2n+1}}{(-2)^{2n+2}}
- 2^{2n+2} (2n+2) \frac{(-1)^{2n+2}}{(-2)^{2n+3}}
\\ = -(2n+2) + (n+1) = - (n+1).$$
Therefore the integral is $$n+1$$ as claimed.

Addendum. We also  need to verify that the residue  at infinity is
zero. We obtain
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{2n+2}
\frac{(1+z)^{2n+2}}{z^{2n+2}} \frac{1}{(2+1/z)^2}
\\ = -\mathrm{Res}_{z=0} (1+z)^{2n+2} \frac{1}{(2z+1)^2} = 0.$$
