If $C_0,C_1,C_2,...,C_n$ are the combinatorial coefficients in the expansion of $(1+x)^n$ then prove that:
$$1C_0^2+3C_1^2+5C_3^2+...+(2n+1)C_n^2=\dfrac{(n+1)(2n)!}{n!n!}=(n+1)\binom {2n}n$$
I am able to compute the linear addition but not the squares of coefficients.
Thanks!
$$\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2=\sum_{k=0}^{n}(2k+1)\binom{n}{k}\binom{n}{n-k}=[x^k]\left[(1+x)^n\sum_{k=0}^{n}\binom{n}{k}(2k+1)x^k\right] $$ but by setting $x=z^2$ we have: $$\sum_{k=0}^{n}\binom{n}{k}(2k+1)x^k=\frac{d}{dz}\sum_{k=0}^{n}\binom{n}{k}z^{2k+1}=(1+x)^{n-1}(1+(2n+1)x)$$ hence: $$\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2= [x^n]\left[(1+x)^{2n}+2nx(1+x)^{2n-1}\right]=\binom{2n}{n}+2n\binom{2n-1}{n}$$ and simplifying: $$\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2=(n+1)\binom{2n}{n} $$ as wanted.
$$\begin{align*} \sum_{k=0}^n(2k+1)\binom{n}k^2&=2\sum_{k=0}^nk\binom{n}k\binom{n}{n-k}+\sum_{k=0}^n\binom{n}k\binom{n}{n-k}\\ &\overset{(1)}=2n\sum_{k=0}^n\binom{n-1}{k-1}\binom{n}{n-k}+\binom{2n}n\\ &=2n\sum_{k=0}^n\binom{n-1}{k-1}\binom{n}{n-k}+\binom{2n}n\\ &\overset{(2)}=2n\binom{2n-1}{n-1}+\binom{2n}n\\ &\overset{(3)}=n\binom{2n}n+\binom{2n}n\\ &=(n+1)\binom{2n}n \end{align*}$$
$(1)$ uses the identity $k\binom{n}k=n\binom{n-1}{k-1}$ and Vandermonde’s identity; $(2)$ is another application of Vandermonde’s identity, and $(3)$ is another application of $k\binom{n}k=n\binom{n-1}{k-1}$.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{n}\pars{2k + 1}{n \choose k}^{2} & = {1 \over 2}\sum_{k = 0}^{n}\braces{\pars{2k + 1}{n \choose k}^{2} + \bracks{2\pars{n - k} + 1}{n \choose n - k}^{2}} \\[5mm] & = \pars{n + 1}\sum_{k = 0}^{n}{n \choose k}^{2} = \bbx{\ds{\pars{n + 1}{2n \choose n}}} \end{align}
$\ds{\sum_{k = 0}^{n}{n \choose k}^{2} = {2n \choose n}}$ is a well known identity. Note that $\ds{\left\lbrace \begin{array}{rcl} \ds{n \choose k} & \ds{=} & \ds{n \choose n - k} \\[2mm] \ds{\sum_{k = 0}^{n}a_{k}} & \ds{=} & \ds{\sum_{k = 0}^{n}a_{n - k}} \end{array}\right.}$
Note that $C_r=\binom nr$.
It is a well-known result that $\sum_{r=0}^n\binom nr^2=\binom {2n}n$ which can be proven easily using the Vandermonde identity.
For even $n$:
The number elements is odd, index from $0$ to $(\frac n2-1)$, $n$, and from $(\frac n2+1)$ to $n$
Since $\binom nr=\binom n{n-r}$, then $\sum_{r=0}^{\frac n2-1}\binom nr^2=\frac 12 \binom {2n}n$.
The summation in the question can then be written as
$$\begin{aligned} &\begin{array}r 1\binom n0^2 &+3\binom n1 ^2 &+\cdots &+(n-1)\binom n{\frac n2-1}^2\\ & & & & +(n+1) \binom n{\frac n2} ^2\\ +(2n+1)\binom nn^2 &+(2n-1)\binom n{n-1}^2 &+\cdots &+(n+3)\binom n{\frac n2+1}^2 \\ =(2n+2)\binom n0^2 &+(2n+2)\binom n1^2 &+\cdots &+(2n+2)\binom n{\frac n2-1}^2 &+(n+1)\binom n{\frac n2}^2 \\ \end{array}\\ &=(2n+2)\cdot \displaystyle\sum_{r=0}^{\frac n2-1}\binom nr^2+(n+1)\binom n{\frac n2}\\ &=(2n+2)\cdot \frac 12 \left[\displaystyle\sum_{r=0}^{n}\binom nr^2-\binom n{\frac n2}\right]+(n+1)\binom n{\frac n2}\\ &=(n+1)\left[\binom {2n}n-\binom n{\frac n2}\right]+(n+1)\binom n{\frac n2}\\ &=\color{red}{(n+1)\binom {2n}n} \end{aligned}$$
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For odd $n$:
The number elements is even, index from $0$ to $\frac {n-1}2$, and from $\frac {n+1}2$ to $n$
Since $\binom nr=\binom n{n-r}$, then $\sum_{r=0}^\frac {n-1}2 \binom nr^2=\frac 12 \binom {2n}n$.
The summation in the question can then be written as
$$\begin{aligned} &\begin{array}r 1\binom n0^2 &+3\binom n1 ^2 &+\cdots &+n\binom n{\frac {n-1}2}^2\\ +(2n+1)\binom nn^2 &+(2n-1)\binom n{n-1}^2 &+\cdots &+(n+2)\binom n{\frac {n+1}2}^2 \\ =(2n+2)\binom n0^2 &+(2n+2)\binom n1^2 &+\cdots &+(2n+2)\binom n{\frac{n-1}2}^2\\ \end{array}\\ &=(2n+2)\displaystyle\sum_{r=0}^{\frac {n-1}2}\binom nr^2\\ &=(2n+2)\cdot \frac 12\displaystyle\sum_{r=0}^{n}\binom nr^2\\ &=(2n+2)\cdot \frac 12 \binom {2n}n\\ &=\color{red}{(n+1)\binom {2n}n} \end{aligned}$$