Prove this sum $\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$ Show that
$$\sum_{k=0}^{n}(-1)^k\cdot 2^{2n-2k}\binom{2n-k+1}{k}=n+1$$

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
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{2n+2}
\frac{(1+z)^{2n+2}}{z^{2n+2}} \frac{1}{(2+1/z)^2}
\\ = -\mathrm{Res}_{z=0} (1+z)^{2n+2} \frac{1}{(2z+1)^2} = 0.$$
