<s>[INST] 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!
 [/INST] The striking symmetry of this binomial identity might lead us to ask what are the roles of $2$ and $-1$ and are there similar identities like this one? Here is a slight generalisation using essentially the same approach as @MarkoRiedel did in his nice answer.

We consider $a\in\mathbb{C}$ instead of $2$ resp. $-1$ and obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n}&\color{blue}{\binom{n+k}{n}a^k}\tag{0}\\
&=\sum_{k=0}^n\binom{-n-1}{k}(-a)^k\tag{1}\\
&=\sum_{k=0}^n[z^k]\frac{1}{(1-az)^{n+1}}\tag{2}\\
&=[z^0]\frac{1}{(1-az)^{n+1}}\sum_{k=0}^n\frac{1}{z^k}\tag{3}\\
&=[z^0]\frac{1}{(1-az)^{n+1}}\cdot\frac{1-z^{n+1}}{z^n(1-z)}\tag{4}\\
&=[z^n]\frac{1}{(1-az)^{n+1}(1-z)}\tag{5}\\
&=\frac{(-1)^n}{a^{n+1}}\mathrm{Res}_{z=0}\underbrace{\frac{1}{z^{n+1}\left(z-\frac{1}{a}\right)^{n+1}(z-1)}}_{f(z)}\tag{6}\\
&=\left(-\frac{1}{a}\right)^{n+1}
\left(\mathrm{Res}_{z=\frac{1}{a}}+\mathrm{Res}_{z=1}+\mathrm{Res}_{z=\infty}\right)f(z)\tag{7}
\end{align*} 

Comment:


*

*In (1) we use $\binom{p}{q}=\binom{p}{p-q}$ and $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (2) and (3) we apply the coefficient of operator and use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (4) we use the finite geometric series formula.

*In (5) we apply the same rule as we did in (3) and skip the term $z^{n+1}$ since it does not contribute to $[z^n]$.

*In (6) we use the representation with residual at $z=0$.

*In (7) we use the fact that computing the residue at zero is minus the sum of the residuals at the other poles and infinity.
Calculating the residuals gives
\begin{align*}
\mathrm{Res}_{z=\frac{1}{a}}f(z)&=\frac{1}{n!}\left[\frac{d^n}{dz^n}\left(\frac{1}{z^{n+1}(z-1)}\right)\right]_{z=\frac{1}{a}}\\
&=\frac{1}{n!}\left[\sum_{q=0}^n\binom{n}{q}\frac{d^q}{dz^q}\left(\frac{1}{z^{n+1}}\right)\frac{d^{n-q}}{dz^{n-q}}\left(\frac{1}{z-1}\right)\right]_{z=\frac{1}{a}}\\
&=\frac{1}{n!}\left[\sum_{q=0}^n\binom{n}{q}(-1)^q\frac{(n+q)!}{q!}\frac{1}{z^{n+1+q}}
(-1)^{n-q}(n-q)!\frac{1}{(z-1)^{n-q+1}}\right]_{z=\frac{1}{a}}\\
&=(-1)^n\left(\frac{a^2}{1-a}\right)^{n+1}\sum_{q=0}^n\binom{n+q}{q}(1-a)^q\\
\mathrm{Res}_{z=1}f(z)&=\frac{1}{\left(1-\frac{1}{a}\right)^{n+1}}\\
&=\left(\frac{a}{a-1}\right)^{n+1}\\
\mathrm{Res}_{z=\infty}f(z)&=\mathrm{Res}_{z=0}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)\\
&=-\mathrm{Res}_{z=0}\left(\frac{1}{z^2}\cdot\frac{z^{n+1}}{\left(\frac{1}{z}-\frac{1}{a}\right)^{n+1}\left(\frac{1}{z}-a\right)}\right)\\
&=-\mathrm{Res}_{z=0}\frac{z^{2n+1}a^{n+1}}{(a-z)^{n+1}(1-z)}\\
&=0
\end{align*}

Putting all together in (7) gives
  \begin{align*}
&\left(-\frac{1}{a}\right)^{n+1}
\left(\mathrm{Res}_{z=\frac{1}{a}}+\mathrm{Res}_{z=1}+\mathrm{Res}_{z=\infty}\right)f(z)\\
&\quad=\left(-\frac{1}{a}\right)^{n+1}\left((-1)^n\left(\frac{a^2}{1-a}\right)^{n+1}\sum_{k=0}^n\binom{n+k}{k}(1-a)^k
+\left(\frac{a}{a-1}\right)^{n+1}+0\right)\\
&\quad\color{blue}{=-\left(\frac{a}{1-a}\right)^{n+1}\sum_{k=0}^n\binom{n+k}{n}(1-a)^k+\left(\frac{1}{1-a}\right)^{n+1}}\tag{8}
\end{align*}

We finally conclude from (0) and (8)

The following is valid for non-negative integer $n$ and $a\in\mathbb{C}$
  \begin{align*}
\color{blue}{(1-a)^{n+1}\sum_{k=0}^n\binom{n+k}{n}a^k
=1-a^{n+1}\sum_{k=0}^n\binom{n+k}{n}(1-a)^k}
\end{align*}
Some special cases:
$a=-1, a=2$ (OP's identity):
\begin{align*}
2^{n+1}\sum_{k=0}^n\binom{n+k}{n}(-1)^k=1+(-1)^n\sum_{k=0}^n\binom{n+k}{n}2^k
\end{align*}
$a=\frac{1}{2}$:
\begin{align*}
\sum_{k=0}^n\binom{n+k}{k}\frac{1}{2^k}=2^n
\end{align*}
$a=\frac{1}{3}, a=\frac{2}{3}$:
\begin{align*}
\left(\frac{2}{3}\right)\sum_{k=0}^n\binom{n+k}{n}\frac{1}{3^k}
=1-\frac{1}{3^{n+1}}\sum_{k=0}^n\binom{n+k}{n}\left(\frac{2}{3}\right)^k
\end{align*}

</s>