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

Question

Recently, a similar summation has been asked where $(k+1)^2$ in denominator could be absorbed in re-making the square of modified binomial coefficient. The binomial summation: $$\sum_{k=0}^{n}\frac{{n\choose k}^2}{k+1}$$ may require some other approach. So the question is how to find this summation by hand, the Mathematica does give an analytic answer.

Answer 1

Since $\binom{n+1}{k+1} = \frac{n+1}{k+1} \binom nk$, we can rewrite this sum as $$\frac1{n+1} \sum_{k=0}^n \binom nk \binom{n+1}{k+1} = \frac1{n+1} \sum_{k=0}^n \binom nk \binom{n+1}{n-k}.$$ By Vandermonde's identity, the sum here simplifies to $\binom{n + (n+1)}{n}$, so the original sum is equal to $\frac1{n+1}\binom{2n+1}{n}$.

Answer 2

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