Help with a Binomial Identity: $\sum_{k=0}^{\infty} \binom{n+k}{k}2^{-k} = 2^{n+1}$ The following is a problem from the 5th edition of Niven's An Introduction to the Theory of Numbers:
Problem 23 of Section 1.4 asks us to prove that

$$\sum_{k=0}^{\infty} \binom{n+k}{k}2^{-k} = 2^{n+1}.$$

I believe I proved it using generating functions, but I would love to have my proof verified and if possible, could someone provide a hint to an alternative proof of the fact. Generating functions have not been covered in this section yet, so I would prefer to figure out how to prove this identity with slightly less powerful tools (if that makes sense).
PROOF The right hand side is the coefficient $[x^n]$ of the generating function $$\frac{2}{1-2x}.$$
On the left hand side, we have that the corresponding generating function of the sequence is
\begin{align*}
\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}\binom{k+n}{n}\frac{1}{2^k}x^n &= \sum_{k=0}^{\infty}\frac{1}{2^k}\sum_{n=0}^{\infty}\binom{n+k}{n}x^n\\
&=\sum_{k=0}^{\infty}\frac{1}{2^k}\frac{1}{(1-x)^{k+1}}\\
&= \frac{1}{1-x}\sum_{k=0}^{\infty}\frac{1}{(2-2x)^k}\\
&= \frac{1}{1-x}\frac{1}{1-\frac{1}{2-2x}}\\
&=\frac{2}{1-2x} = RHS
\end{align*}
 A: An alternative proof is to use a simple induction on $n$. You have
$$x = \sum_{k=0}^\infty \binom{n+1+k}{k}2^{-k}=1+\sum_{k=1}^\infty\binom{n+1+k}{k}2^{-k}=1+\sum_{k=1}^\infty \binom{n+k}{k}2^{-k}+\sum_{k=0}^\infty \binom{n+k+1}{k}2^{-k-1}=\sum_{k=0}^\infty\binom{n+k}{k}2^{-k}+\frac{x}{2}$$
so that applying the IH and solving leads to
$x=2\cdot 2^{n+1}=2^{n+2}$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{k = 0}^{\infty}{n+k \choose  k}2^{-k} = 2^{n+1}:\ {\large ?}}$

\begin{align}&\color{#66f}{\large\sum_{k = 0}^{\infty}{n+k \choose  k}2^{-k}}
=\sum_{k = 0}^{\infty}\pars{-1}^{k}{-n - 1 \choose  k}2^{-k}
=\sum_{k = 0}^{\infty}{-n - 1 \choose  k}\pars{-\,\half}^{k}
\\[3mm]&=\bracks{1 + \pars{-\,\half}}^{-n - 1}=\pars{\half}^{-n - 1}
=\color{#66f}{\Large 2^{n + 1}}
\end{align}

A: This is the same proof provided by @mathse above, expanded for more detail.

