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: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$

One way to look at this is to take:
$\begin{align*}
  \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2
    &= \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k} \binom{n}{n - k} \\
    &= [z^n] \left(
                \sum_k \frac{1}{k + 1} \binom{n}{k} z^k
             \right)
               \cdot (-1)^n \left(
                              \sum_k (-1)^k \binom{n}{k} z^k
                            \right)
\end{align*}$
For the pieces, you know that:
$\begin{align*}
   \sum_k \binom{n}{k} z^k
     &= (1 + z)^n \\
   \sum_k \frac{1}{k + 1} \binom{n}{k} z^k
     &= \frac{1}{z} \sum_k \frac{1}{k + 1} \binom{n}{k} z^{k + 1} \\
     &= \frac{1}{z} \int_0^z (1 + t)^n \, d t \\
     &= \frac{1}{z} \frac{(1 + z)^{n + 1} - 1}{n + 1}
\end{align*}$
Thus:
$\begin{align*}
  \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2
    &= (-1)^n [z^n] \frac{(1 + z)^{n + 1} - 1}{z (n + 1)} (1 - z)^n \\
    &= \frac{(-1)^n}{n + 1}
         [z^{n + 1}] ((1 + z)^{n + 1} - 1) (1 - z)^n \\
    &= \frac{(-1)^n}{n + 1}
         [z^{n + 1}] ((1 - z^2)^n (1 + z) \\
    &= \frac{(-1)^n}{n + 1}
         \left(
           [z^n] (1 - z^2)^n
           [z^{n - 1}] (1 - z^2)^n
         \right) \\
    &= \frac{(-1)^n}{n + 1}
         \left(
            (-1)^{n} \binom{2 n}{n}
              + (-1)^{n - 1} \binom{2 n}{n - 1}
         \right) \\
    &= \frac{1}{n + 1} \left( \binom{2 n}{n} - \binom{2 n}{n - 1} \right) \\
    &= \frac{1}{(n + 1)^2} \binom{2 n}{n}
\end{align*}$
