Binomial identity $\sum_{r=0}^n(2r+1)\binom{n}{r}^2=(n+1)\binom{2n}{n}$

Question

Problem: prove that $\sum_{r=0}^n(2r+1)\binom{n}{r}^2=(n+1)\binom{2n}{n}$

My attempt:

$$ \begin{align} \sum_{r=0}^n(2r+1)\binom{n}{r}^2 &= 2\sum_{r=0}^nr\binom{n}{r}^2 + \sum_{r=0}^n\binom{n}{r}^2 \\ &= 2\sum_{r=1}^n\left[r\cdot \frac{n}{r} \binom{n-1}{r-1}\binom{n}{r}\right] + \binom{2n}{n} \\ &= 2n\sum_{r=1}^n\left[\binom{n-1}{r-1}\binom{n}{r}\right] + \binom{2n}{n} \\ \end{align} $$

Answer 1

You're merely missing $$ \sum_{r=1}^n\binom{n-1}{r-1}\binom{n}{r} = \sum_{r=1}^n\binom{n-1}{r-1}\binom{n}{n-r}=\binom{2n-1}{n-1}=\frac12 \binom{2n}{n}. $$