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

Let
$$
S_n:=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}
$$
Obviously $S_0=1$. Assume that equality
$$S_{n-1}=1\tag1$$
is valid for some $n$. Then it is valid for $n+1$ as well:
$$
\begin{align}
S_n
&=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}\\
&=\sum_{k=0}^n\frac{\binom{n+k-1}{k-1}+\binom{n-1+k}{k}}{2^{n+k}}\\
&=\frac12\sum_{k=0}^{n-1}\frac{\binom{n+k}{k}}{2^{n+k}}+\frac12\sum_{k=0}^n\frac{\binom{n-1+k}{k}}{2^{n-1+k}}\\
&=\frac12S_n-\frac{\binom{2n}{n}}{2^{2n+1}}+\frac12S_{n-1}+\frac{\binom{2n-1}{n}}{2^{2n}}\\
&=\frac12S_n+\frac12S_{n-1}\implies S_n=S_{n-1}\stackrel{I.H.}=1.
\end{align}
$$
Thus, by induction the equality $(1)$ is valid for all integer $n\ge0$.
Accordingly your sum is $2^{2n}S=4^n$.
