A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$

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.

Use Binomial identity: $$(1+t)^n=\sum_{k=0}^{n} {n \choose k}t^n~~~(1)$$ Integration of (1) from $t=0$ to $t=x$,gives $$\frac{(1+x)^{n+1}-1}{n+1}= \sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}~~~(2)$$ Let $t=-1/x$ in (1), then $$(-1)^n x^{-n} (1-x)^n=\sum_{k=0}^{n} (-1)^k {n \choose k} x^{-k}~~~~(3)$$ Multiplying (2) and (3) and collecting the terms of $x^1$, we get $$\frac{(-1)^n}{n+1} [(1-x^2)^{n}(1+x)-(1-x)^n]= x^n\sum_{k=0}^{n} \frac{(-1)^k {n \choose k}^2}{k+1} x^1+...+...$$ $$\implies S_n=\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}=[x^{n+1}] \left((-1)^n \frac{ (1-x^2)^{n}(1+x)-(1-x)^n}{n+1}\right)$$ if $m=n/2]$, then $$S_n=(-1)^{m} \frac{{n \choose m}}{n+1}.$$