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

Question

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:

$$\frac{n\binom{n-1}{k}}{k+1}=\frac{n(n-1)!}{(k+1)k!(n-k-1)!}=\frac{n!}{(k+1)!(n-k-1)!}=\binom{n}{k+1}$$

Answer 1

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 $$\sum_{j=0}^{n-1}\binom{n-1}{j}\binom{n}{n-j-1}=\binom{2n-1}{n-1}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 0}^{n - 1}{{n - 1 \choose k}^{2} \over k + 1} = {{2n \choose n} \over 2n}:\ ?}$

\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}

Answer 3

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 -

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

Answer 4

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.

Answer 5

$$\begin{align}\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}^2}{k+1}&=\frac{1}{n}\sum_{k=0}^{n-1}\binom{n}{k+1}\binom{n-1}{k}\\ &=\frac{1}{n^2}\sum_{k=0}^{n-1}(k+1)\binom{n}{k+1}^2\\ &=\frac{1}{n}\binom{2n-1}{n-1}=\frac{1}{2n}\binom{2n}{n} \end{align}$$

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 $a$.