A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$ Mathematica gives  $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}= ~_2F_1[-n,-n;2;-1],$$ where $~_2F_1$ that is Gauss hypergeometric function. Here the question is: Can one find a simpler closed form for this summation. Recently, the absolute summation for this has been discussed at MSE:
A binomial summation: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$

We seek to evaluate
$$\sum_{k=0}^n \frac{(-1)^k}{k+1} {n\choose k}^2.$$
This is
$$\frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose k+1}
= \frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose n-k}
\\ = [z^n] (1+z)^{n+1} \frac{1}{n+1}
\sum_{k=0}^n (-1)^{k} {n\choose k} z^k
\\ = [z^n] (1+z)^{n+1} \frac{1}{n+1} (1-z)^n
= \frac{1}{n+1} [z^n] (1+z) (1-z^2)^n.$$
Now if $n=2m$ we get
$$\frac{1}{n+1} [z^{2m}] (1+z) (1-z^2)^n
= \frac{1}{n+1} [z^{2m}] (1-z^2)^n
\\ = \frac{1}{n+1} [z^{m}] (1-z)^n
= \frac{1}{n+1} (-1)^m {n\choose m}.$$
On the other hand when $n=2m+1$ we find
$$\frac{1}{n+1} [z^{2m+1}] (1+z) (1-z^2)^n
= \frac{1}{n+1} [z^{2m+1}]z  (1-z^2)^n
\\ = \frac{1}{n+1} [z^{2m}] (1-z^2)^n
= \frac{1}{n+1} [z^{m}] (1-z)^n
= \frac{1}{n+1} (-1)^m {n\choose m}.$$
We thus have even or odd the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n+1} (-1)^{\lfloor n/2\rfloor}
{n\choose \lfloor n/2\rfloor}.}$$
The second case could have been done by inspection given the first.
This result matches the comment by @SangchulLee.
