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:
$$\frac{n\binom{n-1}{k}}{k+1}=\frac{n(n-1)!}{(k+1)k!(n-k-1)!}=\frac{n!}{(k+1)!(n-k-1)!}=\binom{n}{k+1}$$
 A: From your work, it is sufficient to prove the following lemma.
Lemma : 
$$\sum_{j=0}^{k}\binom{n}{j}\binom{m}{k-j}=\binom{n+m}{k}$$
Proof for lemma : 
$\binom nj$ is the coefficient of $x^j$ when we expand $(1+x)^n$.
$$(1+x)^n=\binom n0+\binom n1x+\binom n2x^2+\cdots \binom{n}{n-1}x^{n-1}+\binom nnx^n$$
$$(1+x)^m=\binom m0+\binom m1x+\binom m2x^2+\cdots +\binom{m}{m-1}x^{m-1}+\binom mmx^m$$
Now, $\sum_{j=0}^{k}\binom{n}{j}\binom{m}{k-j}$ represents the coefficient of $x^k$ when we expand $(1+x)^n(1+x)^m=(1+x)^{n+m}$, i.e. $\binom{n+m}{k}$. $\blacksquare$
In the lemma, take $n\to n-1, m\to n, k\to n-1$ to have
$$\sum_{j=0}^{n-1}\binom{n-1}{j}\binom{n}{n-j-1}=\binom{2n-1}{n-1}$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\ds{\sum_{k = 0}^{n - 1}{{n - 1 \choose k}^{2} \over k + 1}
     = {{2n \choose n} \over 2n}:\ ?}$

\begin{align}
&\color{#f00}{\sum_{k = 0}^{n - 1}{{n - 1 \choose k}^{2} \over k + 1}} =
\sum_{k = 0}^{n - 1}{n - 1 \choose k}\
\overbrace{{n - 1 \choose n - 1 - k}}^{\ds{n - 1 \choose k}}\
\overbrace{\int_{0}^{1}x^{k}\,\dd x}^{\ds{1 \over k + 1}}
\\[3mm] = &\
\sum_{k = 0}^{n - 1}{n - 1 \choose k}
\int_{0}^{1}x^{k}\
\overbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{n - 1} \over z^{n - k}}
\,{\dd z \over 2\pi\ic}}^{\ds{n - 1 \choose n - 1 - k}}\,\dd x
\\[3mm] = &\
\oint_{\verts{z} = 1}{\pars{1 + z}^{n - 1} \over z^{n}}
\int_{0}^{1}\
\overbrace{\sum_{k = 0}^{n - 1}{n - 1 \choose k}\pars{xz}^{k}}
^{\ds{\pars{1 + xz}^{n - 1}}}\ \,\dd x\,{\dd z \over 2\pi\ic}
\\[3mm] = &\
\oint_{\verts{z} = 1}{\pars{1 + z}^{n - 1} \over z^{n}}
\int_{0}^{1}\pars{1 + xz}^{n - 1}\,\dd x\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z} = 1}{\pars{1 + z}^{n - 1} \over z^{n}}
{\pars{1 + z}^{n} - 1 \over nz}\,{\dd z \over 2\pi\ic}
\\[3mm] = &
{1 \over n}\
\underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{2n - 1} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}}_{\ds{2n - 1 \choose n}}\ -\
{1 \over n}\
\underbrace{\oint_{\verts{z} = 1}{\pars{1 + z}^{n - 1} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}}_{\ds{=\ {n - 1 \choose n}\ =\ 0}} =
{1 \over n}{2n - 1 \choose n} =
{1 \over n}\,{\pars{2n - 1}! \over n!\pars{n - 1}!}
\\[3mm] = &\
{1 \over 2n}\,\
\underbrace{{\pars{2n}\pars{2n - 1}! \over n!\bracks{n\pars{n - 1}!}}}
_{\ds{2n \choose n}} =
\color{#f00}{\ds{2n \choose n} \over \ds{2n}}
\end{align}
A: In the same spirit as @Felix Marin's answer here Combinatorial proof of summation of $\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}$, we have -
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\begin{align}
{\large\sum_{k\ =\ 0}^{n-1}\dfrac{{n-1 \choose k}^{2}}{k+1}}&=
\sum_{k\ =\ 0}^{n-1}\dfrac{{n -1\choose k}}{k+1}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n-1} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}\biggr(1 + z \biggl)^{n-1}~~
\sum_{k\ =\ 0}^{n-1}\dfrac{{n-1 \choose k}}{k+1}\pars{1 \over z}^{k+1}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}\dfrac{(1 + z )^{n-1}}{n}
\biggr(\pars{1 + {1 \over z}}^{n}-1\biggl)\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n-1} \over (n)z^{n }}\,{\dd z \over 2\pi\ic}
={\dfrac{\large{2n-1 \choose n-1}}{n}}
={\dfrac{\large{2n \choose n}}{2n}}
\end{align}
A: 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.
A: $$\begin{align}\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}&=\frac{1}{n}\sum_{k=0}^{n-1}\binom{n}{k+1}\binom{n-1}{k}\\
&=\frac{1}{n^2}\sum_{k=0}^{n-1}(k+1)\binom{n}{k+1}^2\\
&=\frac{1}{n}\binom{2n-1}{n-1}=\frac{1}{2n}\binom{2n}{n}
\end{align}$$
For the last step, we use a combinatorial argument. Consider the number of ways we can pick a subset $A$ of $\{1,\dots,n\}$ and a subset $B$ of $\{n+1,\dots, 2n\}$ such that $|A|+|B|=n$ and one element of $A$ colored blue.
if $|A|$ has size $k+1$, there are $(k+1)\binom{n}{k+1}^2$ ways to do this. Summing this over $k$ gives the total number.
We can also number of ways to do this by choosing an element $a\in\{1,\dots, n\}$ and coloring it blue and then choosing a subset of size $n-1$ in $\{1,\dots, 2n\}$, not containing $a$.
