Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$
I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me

Inspired by robjohn's answer I show a more general statement: for $x\in \mathbb{R}$ and $n\in\mathbb{N}$,
$$\bbox[5px,border:2px solid #0000A0]{\sum_{k=0}^{n}\binom{n+k}{k}((1-x)^{n+1}x^k+x^{n+1}(1-x)^k)=1}$$
Such identity appears as equation 69 at this page (see hypergeometric's comment).  OP's identity follows by setting $x=1/2$.
Let 
$$f_n(x)=\sum_{k=0}^{n}\binom{n+k}{k}x^k$$
then we have to show by induction that
$$(1-x)^{n+1}f_n(x)+x^{n+1}f_n(x)=1.$$
Base case: for $n=0$  it is trivial. 
\noindent Inductive step step. We note that
\begin{align*}
f_{n+1}(x)
&=\sum_{k=0}^{n+1}\left(\binom{n+k}{k}+\binom{n+1+k-1}{k-1}\right)x^k\\
&=f_n(x)+\binom{2n+1}{n+1}x^{n+1}+xf_{n+1}(x)-\binom{2n+2}{n+1}x^{n+2}\\
&=f_n(x)+\binom{2n+1}{n}x^{n+1}+xf_{n+1}(x)-2\binom{2n+1}{n}x^{n+2}
\end{align*}
Therefore
$$(1-x)f_{n+1}(x)=f_n(x)+\binom{2n+1}{n}x^{n+1}(1-2x)$$
Hence
$$(1-x)^{n+2}f_{n+1}(x)
=(1-x)^{n+1}f_n(x)+\binom{2n+1}{n}(1-x)^{n+1}x^{n+1}(1-2x)$$
and, by replacing $x$ with $(1-x)$, we get
$$x^{n+2}f_{n+1}(1-x)
=x^{n+1}f_n(1-x)-\binom{2n+1}{n}x^{n+1}(1-x)^{n+1}(1-2x).$$
Finally, by adding the last two equations we find
$$(1-x)^{n+2}f_{n+1}(x)+x^{n+2}f_{n+1}(1-x)=
x^{n+1}f_n(1-x)+(1-x)^{n+1}f_n(x)=1$$
and we are done.
