<s>[INST] For all positive integer $n$ prove the equality: $\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}=\frac{\binom{2n}{n}}{2n}$ 
For all positive integer $n$ prove the equality:
$$\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}=\frac{\binom{2n}{n}}{2n}$$

My work so far:
$$\frac{n\binom{n-1}{k}}{k+1}=\frac{n(n-1)!}{(k+1)k!(n-k-1)!}=\frac{n!}{(k+1)!(n-k-1)!}=\binom{n}{k+1}$$
 [/INST] Another approach.
$$ \sum_{k=0}^{n-1}\binom{n-1}{k}^2\frac{1}{k+1}=\int_{0}^{1}\color{blue}{\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n-1}{n-1-k}x^k}\,dx \tag{1}$$
and the blue term is the coefficient of $y^{n-1}$ in the following product:
$$\left(\sum_{k=0}^{n-1}\binom{n-1}{k}x^k y^k\right)\cdot\left(\sum_{k=0}^{n-1}\binom{n-1}{k}y^k\right)=(1+xy)^{n-1}(1+y)^{n-1}\tag{2}$$
so the original sum is the coefficient of $y^{n-1}$ in:
$$ (1+y)^{n-1}\int_{0}^{1}(1+xy)^{n-1}\,dx =(1+y)^{n-1}\cdot\frac{-1+(1+y)^{n}}{ny}\tag{3}$$
or $\frac{1}{n}$ times the coefficient of $y^n$ in $(1+y)^{2n-1}-(1+y)^{n-1}$. That leads to:

$$ \sum_{k=0}^{n-1}\binom{n-1}{k}^2\frac{1}{k+1}=\frac{1}{n}\binom{2n-1}{n}=\color{red}{\frac{1}{2n}\binom{2n}{n}}\tag{4}$$

as wanted.
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