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 $S_{2}$

We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^n}\color{blue}{(-4)^k\binom{n+k}{2k}}&=\sum_{k=0}^n(-4)^k\binom{n+k}{n-k}\\
&=\sum_{k=0}^n(-4)^k[z^{n-k}](1+z)^{n+k}\tag{2}\\
&=[z^n](1+z)^n\sum_{k=0}^n(-4z(1+z))^k\tag{3}\\
&=[z^n](1+z)^n\frac{1-(-4z(1+z))^{n+1}}{1+4z(1+z)}\tag{4}\\
&=[z^n](1+z)^n\frac{1}{1+4z(1+z)}\tag{5}\\
&=[z^n]\frac{(1+z)^n}{(1+2z)^2}\\
&=[z^n](1+z)^n\sum_{j=0}^\infty\binom{-2}{j}(2z)^j\tag{6}\\
&=[z^n](1+z)^n\sum_{j=0}^\infty(j+1)(-2z)^j\tag{7}\\
&=\sum_{j=0}^n(j+1)(-2)^j[z^{n-j}](1+z)^n\\
&=\sum_{j=0}^n(j+1)(-2)^j\binom{n}{n-j}\tag{8}\\
&=\sum_{j=1}^n\binom{n}{j}j(-2)^j+\sum_{j=0}^n\binom{n}{j}(-2)^j\\
&=n\sum_{j=1}^n\binom{n-1}{j-1}(-2)^j+(1+(-2))^n\tag{9}\\
&=n\sum_{j=0}^{n-1}\binom{n-1}{j}(-2)^{j+1}+(-1)^n\tag{10}\\
&=(-2n)(1+(-2))^{n-1}+(-1)^n\\
&\,\,\color{blue}{=(-1)^n(2n+1)}
\end{align*}

Comment:


*

*In (2) we apply the coefficient of operator according to (1).

*In (3) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (4) we apply the finite geometric series formula.

*In (5) we skip the term which does not contribute to $[z^n]$.

*In (6) we apply the binomial series expansion.

*In (7) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (8) we select the coefficient of $z^{n-j}$.

*In (9) we apply the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

*In (10) we shift the index to start with $j=0$.
