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!

Here is a proof of the general case below. My original answer is at the end.

$$\forall m,n\in\Bbb N,\ \forall x\in\Bbb R,\ \sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}$$

Let $m,n\in\Bbb N$, $f:x\mapsto\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^kx^{n+1}$, we have $f(0)=0$ and for all $x\in\Bbb R$,
\begin{align}
f'(x)&=(n+1)x^{n}\sum_{k=0}^{m}\binom{n+k}{k}(1-x)^k-x^{n+1}\sum_{k=0}^{m}\binom{n+k}{k}k(1-x)^{k-1}\\
&=x^{n}(1-x)^{m}\cdot A
\end{align}
where
\begin{align}
A&=\sum_{k=0}^{m}\binom{n+k}{k}(n+1)(1-x)^{k-m}-\sum_{k=0}^{m}\binom{n+k}{k}kx(1-x)^{k-m-1}\\
&=\sum_{k=0}^{m}\binom{n+k}{k}(n+k+1)(1-x)^{k-m}-\sum_{k=0}^{m}\binom{n+k}{k}k(1-x)^{k-m-1}\tag1\\
&=\sum_{k=1}^{m+1}\binom{n+k}{k}k(1-x)^{k-m-1}-\sum_{k=1}^{m}\binom{n+k}{k}k(1-x)^{k-m-1}\tag2\\
&=\binom{n+m+1}{m+1}(m+1)\\
&=\frac{(n+m+1)!}{n!\ m!}
\end{align}
$(1):-x(1-x)^{k-m-1}=(1-\frac1{1-x})(1-x)^{k-m}=(1-x)^{k-m}-(1-x)^{k-m-1}$
$(2):\binom{n+k}{k}(n+k+1)=\frac{(n+k+1)!}{n!k!}=\binom{n+k+1}{k+1}(k+1)$
Thus,
\begin{align}
\forall x\in \Bbb R,\ f(x)&=A\int_0^xt^n(1-t)^mdt\\
&=A\int_{1-x}^1(1-s)^ns^mds\tag{$s=1-t$}\\
&=A\int_0^1(1-s)^ns^mds-A\int_0^{1-x}(1-s)^ns^mds\\
&=1-\sum_{k=0}^{n}\binom{m+k}{k}x^k(1-x)^{m+1}\tag3
\end{align}
$(3):$ integration by parts proves that $\int_0^1(1-s)^ns^mds=\frac{n!\ m!}{(n+m+1)!}=A^{-1}$, and $m\leftrightarrow n$ and $x\leftrightarrow 1-x$ in the expressions of $f$ give the second term.

Original answer:
The general formula of your question is $I_x(a,b)=1-I_{1-x}(b,a)$
  where $I$ is the regularized incomplete beta function. The
  negative binomial distribution may also be helpful.
Note that $\forall n\in\Bbb N,\ \forall x\in\Bbb R,\
I_x(n+1,n+1)=\sum^{n}_{k=0}\binom{n+k}{k}(1-x)^kx^{n+1}$.
The question is to prove that
  $$\sum_{k=0}^{n}{{n+k}\choose{n}}2^k(-1)^{n+1}=1-\sum_{k=0}^{n}{{n+k}\choose{n}}(-1)^{k}2^{n+1}$$
  which is a special case for $I_{-1}(n+1,n+1)=1-I_2(n+1,n+1)$.

