1

Let $n \in \Bbb N$ be fixed. Let $C_r = \binom n r\ $ for $0 \leq r \leq n.$ Evaluate $$C_0^2 + 3 C_1^2 + \cdots + (2n+1) C_n^2.$$

Any hint in this regard will be highly appreciated. Thank you very much.

math maniac.
  • 1,993

2 Answers2

3

Using $k\binom{n}{k}=n\binom{n-1}{k-1}$ for $k>0$, and Chu-Vandermonde identity, \begin{align}\sum_{k=0}^{n}(2k+1)\binom{n}{k}^2&=2n\sum_{k=1}^{n}\binom{n-1}{k-1}\binom{n}{k}+\sum_{k=0}^{n}\binom{n}{k}^2\\&=2n\sum_{k=0}^{n-1}\binom{n-1}{k}\binom{n}{n-1-k}+\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}\\&=2n\binom{2n-1}{n-1}+\binom{2n}{n}=\color{blue}{(n+1)\binom{2n}{n}}.\end{align}

metamorphy
  • 39,111
  • Oh! Brilliant. Nicely explained! +1. – math maniac. Jan 21 '20 at 08:23
  • How did you come up with such a nice answer? Trust me! I am pretty much impressed by your solution. Love you! I am very eager to know your thought process behind it. Thank you so much. – math maniac. Jan 21 '20 at 08:25
  • Well, this works for any sum of the form $\sum_{k=0}^{n}\binom{n}{k}^2 p(k)$, where $p(k)$ is a polynomial in $k$ (just represent $p(k)$ as $p_0+p_1 k+p_2 k(k-1)+\ldots$). An alternative solution is to recall that $\sum\limits_{k=0}^{n}a_{n-k}b_k$ is the coefficient of $x^n$ in $\left(\sum\limits_{k=0}^{n}a_k x^k\right)\left(\sum\limits_{k=0}^{n}b_k x^k\right)$, and put $a_k=\binom{n}{k}$ and $b_k=\binom{n}{k}p(k)$. – metamorphy Jan 21 '20 at 08:33
  • I do not know where I am wrong since I arrive at $\color{blue}{(2n+1)\binom{2n}{n}}$ – Claude Leibovici Jan 21 '20 at 09:18
  • @ClaudeLeibovici: $n=1$ is easy to check (and track back)... – metamorphy Jan 21 '20 at 09:34
0

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

\begin{align} &\bbox[5px,#ffd]{\sum_{r = 0}^{n}\pars{2r + 1}{n \choose r}^{2}} = \sum_{r = 0}^{n}\pars{2r + 1}{n \choose r}{n \choose n - r} \\[5mm] = &\ \sum_{r = 0}^{n}\pars{2r + 1}{n \choose r}\bracks{z^{n - r}} \pars{1 + z}^{n} = \bracks{z^{n}} \pars{1 + z}^{n}\sum_{r = 0}^{n}z^{r}\pars{2r + 1}{n \choose r} \\[5mm] = &\ \bracks{z^{n}} \pars{1 + z}^{n}\pars{2z\,\partiald{}{z} + 1}\sum_{r = 0}^{n}{n \choose r}z^{r} \\[5mm] = &\ \bracks{z^{n}} \pars{1 + z}^{n}\pars{2z\,\partiald{}{z} + 1}\pars{1 + z}^{n} \\[5mm] = &\ \bracks{z^{n}} \pars{1 + z}^{n}\bracks{2nz\pars{1 + z}^{n - 1} + \pars{1 + z}^{n}} \\[5mm] = &\ 2n\bracks{z^{n - 1}}\pars{1 + z}^{2n - 1} + \bracks{z^{n}}\pars{1 + z}^{2n} = 2n{2n - 1 \choose n - 1} + {2n \choose n} \\[5mm] = &\ n\ \underbrace{\bracks{{2n \over n}\,{\pars{2n - 1}! \over \pars{n - 1}!\, n!}}}_{\ds{2n \choose n}}\ +\ {2n \choose n} \\[5mm] = &\ \bbx{\pars{n + 1}{2n \choose n}} \\ & \end{align}
Felix Marin
  • 89,464