<s>[INST] Identify $\sum(-1)^{n-k}2^{2k}\binom{n+k}{2k}$ Does anybody know what the following sum evaluates to?
$$
\sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k}
$$
 [/INST] Generating functions to the rescue again! Defining
$$
c_n = \sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k},
$$
we have
\begin{align*}
\sum_{n=0}^\infty c_n x^n &= \sum_{n=0}^\infty x^n \sum_{k=0}^n{(-1)^{n-k}}2^{2k}\binom{n+k}{2k} \\
&= \sum_{k=0}^\infty {(-1)^{k}}2^{2k} \sum_{n=k}^\infty (-x)^n \binom{n+k}{2k} \\
&= \sum_{k=0}^\infty {(-1)^{k}}2^{2k} \frac{(-x)^{k}}{(1-(-x))^{2k+1}} \\
&= \frac1{1+x} \sum_{k=0}^\infty \bigg( \frac{4x}{(1+x)^2} \bigg)^k \\
&= \frac1{1+x} \frac1{1-4x/(1+x)^2} = \frac{1+x}{(1-x)^2};
\end{align*}
and one can verify that
$$
\frac{1+x}{(1-x)^2} = \sum_{n=0}^\infty (2n+1)x^n.
$$
It's true that the middle equality in the long calculation isn't obvious, but
$$
\sum_{n} \binom nj x^n = \frac{x^j}{(1-x)^{j+1}}
$$
is a standard formula in the generating functions biz.
</s>