How to solve this binomial summation Problem where I was stuck at:
$$\sum_{k=0}^{n} (-4)^k\cdot \binom{n+k}{ 2k },$$ what I tried was to get observe that inside term is coefficients in expansion of $(1+2x)^{n+k}$ but evaluating it afterwards is not working for me.

We seek to evaluate
$$\sum_{k=0}^n (-1)^k 2^{2k} {n+k\choose 2k}$$
which is
$$\sum_{k=0}^n (-1)^k 2^{2k} {n+k\choose n-k}
= [z^n] (1+z)^n \sum_{k=0}^n (-1)^k 2^{2k} (1+z)^k z^k.$$
Here the coefficient extractor enforces the upper limit of the sum and
we find
$$[z^n] (1+z)^n \frac{1}{1+4z(1+z)} =
[z^n] (1+z)^n \frac{1}{(1+2z)^2}.$$
This is
$$\frac{1}{4} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} (1+z)^n
\frac{1}{(z+1/2)^2}.$$
Residues sum to zero and the residue at infinity is zero by inspection.
Hence the residue at zero is minus the residue at $-1/2.$ We have
$$\frac{1}{4}
\left.\left( \frac{1}{z^{n+1}} (1+z)^n \right)'\right|_{z=-1/2}
= \frac{1}{4}
\left.\left( - \frac{n+1}{z^{n+2}} (1+z)^n
+ \frac{n}{z^{n+1}} (1+z)^{n-1}\right)\right|_{z=-1/2}
\\ = \frac{1}{4} \left( - (n+1) (-1)^n 2^{n+2} \frac{1}{2^n}
+ n (-1)^{n+1} 2^{n+1} \frac{1}{2^{n-1}} \right)
\\ = -(n+1) (-1)^n + n (-1)^{n+1}.$$
Taking into account the sign we get at last
$$\bbox[5px,border:2px solid #00A000]{
(-1)^n \times (2n+1).}$$
