Prove that $(-1)^n\sum_{k=0}^{n}{{{n+k}\choose{n}}2^k}+1=2^{n+1}\sum_{k=0}^{n}{{{n+k}\choose{n}}(-1)^k}$ Define:
$$A_n:=\sum_{k=0}^{n}{{n+k}\choose{n}} 2^k,\quad{B_n}:=\sum_{k=0}^{n}{{n+k}\choose{n}}(-1)^k$$
I've found that (based on values for small $n$) this identity seems to be true:
$${\left(-1\right)}^nA_n+1=2^{n+1}B_n$$
However, I'm stuck on trying to find a proof. Any ideas? Thanks!

Using formal power series we have the Iverson bracket
$$[[0\le k\le n]] = [z^n] z^k \frac{1}{1-z}.$$
We then get for $A_n$
$$\sum_{k\ge 0} [z^n] z^k \frac{1}{1-z} {n+k\choose k} 2^k
= [z^n] \frac{1}{1-z}
\sum_{k\ge 0} z^k {n+k\choose n} 2^k
\\ = [z^n] \frac{1}{1-z} \frac{1}{(1-2z)^{n+1}}.$$
This yields for $1+(-1)^n A_n$
$$1 + (-1)^n [z^n] \frac{1}{1-z} \frac{1}{(1-2z)^{n+1}}
= 1 + [z^n] \frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}$$
This is
$$1 + \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}.$$
Now the residue at infinity is zero by inspection and we get
the closed form (residues sum to zero)  
$$1 - \mathrm{Res}_{z=-1} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}
- \mathrm{Res}_{z=-1/2} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}
\\ = - \mathrm{Res}_{z=-1/2} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(1+2z)^{n+1}}
\\ = - \frac{1}{2^{n+1}}\mathrm{Res}_{z=-1/2} \frac{1}{z^{n+1}}
\frac{1}{1+z} \frac{1}{(z+1/2)^{n+1}}
.$$
We evidently require (Leibniz rule)
$$\frac{1}{n!} \left(\frac{1}{z^{n+1}} \frac{1}{1+z} \right)^{(n)}
\\ = \frac{1}{n!}
\sum_{q=0}^n {n\choose q} \frac{(-1)^q (n+q)!}{n!}
\frac{1}{z^{n+1+q}}
\frac{(n-q)! (-1)^{n-q}}{(1+z)^{n-q+1}}
\\ = (-1)^n \sum_{q=0}^n {n+q\choose q}
\frac{1}{z^{n+1+q}}
\frac{1}{(1+z)^{n-q+1}}.$$
Evaluate at $z=-1/2$ to get
$$(-1)^n \sum_{q=0}^n {n+q\choose q}
(-2)^{n+1+q} 2^{n-q+1}
= 2^{2n+2} \sum_{q=0}^n {n+q\choose q} (-1)^{q+1}.$$
Restoring the multiplier in front now yields
$$- \frac{1}{2^{n+1}} 2^{2n+2}
\sum_{q=0}^n {n+q\choose q} (-1)^{q+1}
= 2^{n+1} \sum_{q=0}^n {n+q\choose q} (-1)^{q}.$$
This is $2^{n+1} B_n$ as claimed.
