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.

With formal power series as requested by OP we may write
$$\sum_{k=0}^n 2^k {2n-k\choose n}
= 2^n \sum_{k=0}^n 2^{-k} {n+k\choose n}
= 2^n \sum_{k=0}^n 2^{-k} [z^n] (1+z)^{n+k}
\\ = 2^n [z^n] (1+z)^n \sum_{k=0}^n 2^{-k} (1+z)^k
= 2^n [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-(1+z)/2}
\\ = 2^{n+1} [z^n] (1+z)^n \frac{1-(1+z)^{n+1}/2^{n+1}}{1-z}.$$
We get from the first piece
$$2^{n+1} [z^n] (1+z)^n \frac{1}{1-z} = 
2^{n+1} \sum_{k=0}^n {n\choose k} = 2^{2n+1}.$$
The second piece yields
$$- [z^n] (1+z)^{2n+1} \frac{1}{1-z}
= - \sum_{k=0}^n {2n+1\choose k}
= - \frac{1}{2} 2^{2n+1}.$$
Joining the two pieces we find
$$2^{2n+1} - 2^{2n} = 2^{2n} = 4^n.$$
