I came across the following sum to simplify and completely stuck:
$$\sum_{k = 0}^n \frac{{n\choose k}^2}{k+1}$$
I know Vandermonde equation, differentiation and integration methods for binomial coefficients; however, I still don't see a way to solution.
On
Integration of Binomial series $$(1+t)^n=\sum_{k=0}^n {n \choose k} t^k$$ from $t=0$ to $t=x$ gives $$\frac{(1+x)^{n+1}-1}{x(n+1)}=\sum_{k=0}^{n}\frac{{n \choose k}x^{k}}{k+1}. \tag{1}$$ Changing $t$ to $1/x$ in (1) we have $$\left(1+\frac{1}{x}\right)^{n}=\sum_{k=0}^{n}{n \choose k}\left(\frac{1}{x}\right)^k. \tag{2}$$ Multiplying (1) and (2), we get $$\frac{(1+x)^{2n+1}-(1+x)^{n}}{(n+1)}=x^{n+1}\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}+x^{n+1}\sum_{k=1}^{n} A_k x^{k}. \tag{3}$$ The required sum is nothing but the coefficient of $x^{n+1}$ in the L.H.S. of (3). Finally, we get $$\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}=\frac{{2n+1 \choose n}}{n+1}.$$
Absorb the $k+1$ and apply Vandermonde: \begin{align} \sum_{k=0}^n \frac{\binom{n}{k}^2}{k+1} &=\frac{1}{n+1}\sum_{k=0}^n \frac{n+1}{k+1}\binom{n}{k}^2\\ &=\frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{k+1}\binom{n}{k}\\ &=\frac{1}{n+1}\sum_{k=0}^n \binom{n+1}{n-k}\binom{n}{k}\\ &=\frac{1}{n+1}\binom{2n+1}{n} \end{align}