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.
One way to look at this is to take:

$\begin{align*}
  \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2
    &amp;= \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k} \binom{n}{n - k} \\
    &amp;= [z^n] \left(
                \sum_k \frac{1}{k + 1} \binom{n}{k} z^k
             \right)
               \cdot (-1)^n \left(
                              \sum_k (-1)^k \binom{n}{k} z^k
                            \right)
\end{align*}$

For the pieces, you know that:

$\begin{align*}
   \sum_k \binom{n}{k} z^k
     &amp;= (1 + z)^n \\
   \sum_k \frac{1}{k + 1} \binom{n}{k} z^k
     &amp;= \frac{1}{z} \sum_k \frac{1}{k + 1} \binom{n}{k} z^{k + 1} \\
     &amp;= \frac{1}{z} \int_0^z (1 + t)^n \, d t \\
     &amp;= \frac{1}{z} \frac{(1 + z)^{n + 1} - 1}{n + 1}
\end{align*}$

Thus:

$\begin{align*}
  \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2
    &amp;= (-1)^n [z^n] \frac{(1 + z)^{n + 1} - 1}{z (n + 1)} (1 - z)^n \\
    &amp;= \frac{(-1)^n}{n + 1}
         [z^{n + 1}] ((1 + z)^{n + 1} - 1) (1 - z)^n \\
    &amp;= \frac{(-1)^n}{n + 1}
         [z^{n + 1}] ((1 - z^2)^n (1 + z) \\
    &amp;= \frac{(-1)^n}{n + 1}
         \left(
           [z^n] (1 - z^2)^n
           [z^{n - 1}] (1 - z^2)^n
         \right) \\
    &amp;= \frac{(-1)^n}{n + 1}
         \left(
            (-1)^{n} \binom{2 n}{n}
              + (-1)^{n - 1} \binom{2 n}{n - 1}
         \right) \\
    &amp;= \frac{1}{n + 1} \left( \binom{2 n}{n} - \binom{2 n}{n - 1} \right) \\
    &amp;= \frac{1}{(n + 1)^2} \binom{2 n}{n}
\end{align*}$