Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$ 
Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$


This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
Method 2:
$$S=\text{coefficient of $x^n$ in }:$$
$$2^n(1+x)^n\Bigg( 1+\frac{1+x}{2}+\left(\frac{1+x}{2}\right)^2+\cdots+\left(\frac{1+x}{2}\right)^n \Bigg)$$
$$=(1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right)$$
It looks like a heavy task to collect all the $x^n$ coefficients from this expression.

Im out of ideas. Any hint is appreciated.
From Putnam 2020, Q A2

Consider
$$\left( \begin{matrix}
   n  \\
   k  \\
\end{matrix} \right)=\frac{1}{2\pi i}\int\limits_{\left| z \right|=R}^{{}}{\frac{{{\left( 1+z \right)}^{n}}}{{{z}^{k+1}}}dz}$$
so
$$\sum\limits_{r=0}^{n}{{{2}^{n-r}}\left( \begin{matrix}
   n+r  \\
   r  \\
\end{matrix} \right)}=\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\sum\limits_{r=0}^{n}{{{2}^{n-r}}}\frac{{{\left( 1+z \right)}^{n+r}}}{{{z}^{r+1}}}dz}=\\\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\frac{{{2}^{n+1}}{{\left( 1+z \right)}^{n}}}{z-1}-\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( z-1 \right){{z}^{n+1}}}dz}\\=\frac{1}{2\pi i}\int\limits_{\left| z \right|<1}^{{}}{\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( 1-z \right){{z}^{n+1}}}dz}$$
Where for convenience we've chosen a circular contour enclosing the origin with radius less than $1$.  We can do this because directly after the summation the only contribution comes from a residue at $z=0$ (note there is no residue at $z=1$ in the second line).  The final line above reflects this. Now to get the residue at $z=0$...
$$res\frac{{{\left( 1+z \right)}^{1+2n}}}{\left( 1-z \right){{z}^{n+1}}}=res\sum\limits_{m=0}^{\infty }{{}}\sum\limits_{k=0}^{\infty }{\left( \begin{matrix}
   1+2n  \\
   k  \\
\end{matrix} \right)\frac{1}{{{z}^{n+1-m-k}}}}\\=\sum\limits_{m=0}^{\infty }{\left( \begin{matrix}
   1+2n  \\
   n-m  \\
\end{matrix} \right)=}\sum\limits_{m=0}^{n}{\left( \begin{matrix}
   1+2n  \\
   n-m  \\
\end{matrix} \right)}=\sum\limits_{m=0}^{n}{\left( \begin{matrix}
   1+2n  \\
   m  \\
\end{matrix} \right)}={{4}^{n}}$$
hence
$$\sum\limits_{r=0}^{n}{{{2}^{n-r}}\left( \begin{matrix}
   n+r  \\
   r  \\
\end{matrix} \right)}={{4}^{n}}$$
