<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] Let 
$$S_n=\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}\tag{*}$$
This issue can be connected in a natural way to Chebyshev polynomials of the second kind, with explicit expression see Wikipedia article.
$$U_n(x)=\sum_{k=0}^n(-2)^k\binom{n+k+1}{2k+1}(1-x)^k\tag{1}$$
If we take $x=-1$ in (1), 
$$U_n(-1)=\sum_{k=0}^n(-4)^k\binom{n+k+1}{2k+1}=(-1)^n (n+1),\tag{2}$$
the second expression for $U_n(-1)$  being a known property (see Remark below).
Now, let us write the same expression as (2), but for $n-1$ instead of $n$ :
$$U_{n-1}(-1)=\sum_{k=0}^{n-1}(-4)^k\binom{n+k}{2k+1}=(-1)^{n-1} (n)\tag{3}$$
Subtracting (3) from (2), 
$$\sum_{k=0}^{n-1}(-4)^k\underbrace{\left(\binom{n+k+1}{2k+1}-\binom{n+k}{2k+1}\right)}_{\binom{n+k}{2k}}+(-4)^n=\underbrace{(-1)^n (n+1) - (-1)^{n-1} (n)}_{(-1)^n (2n+1)}\tag{4}$$
(we have used the classical recurrence relationship between binomial coefficients)
It is finished there because (4) expresses the fact that $S_n$ (given by (*)) is :

$$S_n=(-1)^n (2n+1)$$

Remark : Why do we have $U_n(-1)=(-1)^n (n+1)$ ?
It can be easily seen by setting $x=-1$ into the generating function of the $U_n$s :
$$\sum_{n=0}^{\infty}U_n(x)t^n=\dfrac{1}{1-2tx+t^2}$$
giving
$$\dfrac{1}{(1+t)^2}=1-2t+3t^2-4t^3+\cdots$$
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