Proving $4^n=\sum_{k=0}^n2^k\binom{2n-k}{n}$ $4^n = \sum\limits_{k=0}^{n}2^k\cdot{{2n - k} \choose n}$
I tried formal power series, but failed.

A solution through Complex Analysis and the residue theorem.
$$\mathcal{S}(n)=\sum_{k=0}^{n} 2^k\binom{2n-k}{n} = \sum_{k=0}^{n}2^{n-k}\binom{n+k}{n} $$
is the coefficient of $x^n$ in the product between $\sum_{k\geq 0}2^k x^k=\frac{1}{1-2x}$ (geometric series) and $\sum_{k\geq 0}\binom{n+k}{n}x^k = \frac{1}{(1-x)^{n+1}}$ (stars and bars). In particular
$$ \mathcal{S}(n)=\text{Res}\left(\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}},x=0\right) $$
but due to the symmetry of the meromorphic function $\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}}$ the residue at $0$ and the residue at $1$ are the same number. The only other pole is at $x=\frac{1}{2}$ and the sum of the residues is zero, hence $\mathcal{S}(n)=4^n$ can be proved from the straightforward
$$ \text{Res}\left(\frac{1}{(1-2x)\left[x(1-x)\right]^{n+1}},x=\frac{1}{2}\right)=-2\cdot 4^n. $$
