<s>[INST] sum of binomial series with alternate terms 
Evaluation of series $\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}$

what I tried:
from Binomial Identity
$$\binom{n+k}{2k}=\binom{n+k-1}{2k}+\binom{n+k-1}{2k-1}$$
series is $$\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k}+\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k-1}$$
let $\displaystyle S_{1}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{0}-4\binom{n}{2}+4^2\binom{n+1}{4}+\cdots +(-4)^n\binom{2n-1}{2n}$
let $\displaystyle S_{2}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{-1}-4\binom{n}{1}+4^2\binom{n+1}{3}+\cdots +(-4)^n\binom{2n-1}{2n-1}$
Help me please
did not know to simplify $S_{1}$ and $S_{2}$
 [/INST] With the following the purpose was to use slightly different functions
from what we saw in the answer by @MarkusScheuer. Start from
$$\sum_{k=0}^n (-4)^k {n+k\choose 2k}
= \sum_{k=0}^n (-4)^k [z^{n-k}] \frac{1}{(1-z)^{2k+1}}
\\ = [z^n] \frac{1}{1-z}
\sum_{k=0}^n (-4)^k \frac{z^k}{(1-z)^{2k}}.$$
Here the coefficient extractor enforces the range and we get
$$[z^n] \frac{1}{1-z}
\sum_{k\ge 0} (-4)^k \frac{z^k}{(1-z)^{2k}}
= [z^n] \frac{1}{1-z} \frac{1}{1+4z/(1-z)^2}
\\ = [z^n] \frac{1-z}{(1-z)^2+4z}
= [z^n] \frac{1-z}{(1+z)^2}
\\ = [z^n] \frac{1}{(1+z)^2} - [z^{n-1}] \frac{1}{(1+z)^2}
\\ = (-1)^n (n+1) - (-1)^{n-1} n = (-1)^n (2n+1).$$
</s>