Find the sum using properties of binomial coefficients

Question

I need to find the following sum using properties of binomial coefficients:$$\sum\limits_{k=0}^{n} k^3 {n\choose k}^2.$$ I transformed this sum first in:$n^2\sum\limits_{k=1}^{n} k{n-1\choose k-1}^2$. Then after substitution $j=k-1$ I get $n^2\sum\limits_{j=0}^{n-1}(j+1){n-1\choose j}^2$. Then, after separating this in two sums for first one I get: $n^2\sum\limits_{j=0}^{n-1}j {n-1\choose j}^2$. And for the second $\sum\limits_{j=0}^{n-1}{n-1 \choose j}^2$. I am not sure how to solve these two. Probably is has something to do with Vandermonde convolution, but honestly, I don't know to use it properly.

Answer 1

Once you reduce it to

$$n^2\sum_{k=0}^{n-1}(k+1)\binom{n-1}k^2\,,\tag{1}$$

you can finish the calculation combinatorially.

Suppose that you have a pool of $n$ women and $n$ men, and you want to form a committee of $n+1$ people. The committee is to have a man and a woman as co-chairs, and it is to have a treasurer, who must be a women and who can be the female co-chair. There are $n^2$ ways to choose the co-chairs. Once that has been done, there are $$\binom{n-1}k\binom{n-1}{n-1-k}=\binom{n-1}k^2$$ ways to choose the remaining $n-1$ members of the committee in such a way that there are altogether $k+1$ women on the committee, and there are $k+1$ ways to choose which woman is to be the treasurer, so the number of different outcomes is given by $(1)$.

Alternatively, after choosing the co-chairs we could choose the treasurer and then the rest of the committee. There are $n^2$ ways to choose the co-chairs. If we decide that the female co-chair is also the treasurer, we can fill out the committee with any $n-1$ of the $2n-2$ remaining people, so there are $n^2\binom{2n-2}{n-1}$ outcomes in which the treasurer is also a co-chair. Otherwise, there are $n-1$ ways to choose one of the remaining women to be the treasurer, after which we can fill out the committee with any $n-2$ of the remaining $2n-3$ people, so there are $n^2(n-1)\binom{2n-3}{n-2}$ outcomes in which the three offices are held by three different people. It follows that

$$n^2\sum_{k=0}^{n-1}(k+1)\binom{n-1}k^2=n^2\binom{2n-2}{n-1}+n^2(n-1)\binom{2n-3}{n-2}\,.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Lets $\ds{{\cal I}_{m} \equiv \sum_{k = 0}^{n}k^{m}{n \choose k}^{2}\,,\qquad {\cal I}_{3}:\ {\Large ?}}$.

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\begin{align} {\cal I}_{3} & = \bbox[5px,#ffd]{\sum_{k = 0}^{n}k^{3}{n \choose k}^{2}} = \sum_{k = 0}^{n}\pars{n - k}^{3}{n \choose n - k}^{2} \\[5mm] = & \sum_{k = 0}^{n}\pars{n^{3} - 3n^{2}k + 3nk^{2} - k^{3}}{n \choose k}^{2} \\[5mm] \implies {\cal I}_{3} & = {1 \over 2}\,n^{3}\,{\cal I}_{0} - {3 \over 2}\,n^{2}\,{\cal I}_{1} + {3 \over 2}\,n \,{\cal I}_{2}\,,\qquad \pars{\substack{\ds{Similarly,} \\[1mm] \ds{{\cal{I}_{1} = {1 \over 2}\,n\,{\cal I}_{0}}}}} \\[5mm] \implies {\cal I}_{3} & = -\,{1 \over 4}\,n^{3}\,{\cal I}_{0} + {3 \over 2}\,n \,{\cal I}_{2} \end{align}

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\begin{align} \sum_{k = 0}^{n}{n \choose k}^{2}x^{k} & = \sum_{k = 0}^{n}{n \choose k}x^{k}{n \choose n - k} \\[5mm] & = \sum_{k = 0}^{n}{n \choose k}x^{k}\bracks{z^{n - k}} \pars{1 + z}^{n} \\[5mm] & = \bracks{z^{n}} \pars{1 + z}^{n}\sum_{k = 0}^{n} {n \choose k}\pars{xz}^{k} \\[5mm] & = \bracks{z^{n}}\pars{1 + z}^{n}\pars{1 + xz}^{n} \end{align}

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$$ {\cal I}_{0} = \bracks{z^{n}}\pars{1 + z}^{2n} = \bbx{2n \choose n} \\ $$

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\begin{align} {\cal I}_{2} & = \sum_{k = 0}^{n}k^{2}{n \choose k}^{2} = \bracks{z^{n}} \pars{1 + z}^{n}\, \left.\partiald{}{x}\pars{x\,\partiald{}{x}} \pars{1 + xz}^{n}\,\right\vert_{\, x\ =\ 1} \\[5mm] & = n\bracks{z^{n - 1}} \pars{1 + z}^{n}\, \left.\partiald{\bracks{x\pars{1 + xz}^{n - 1}}}{x} \,\right\vert_{\, x\ =\ 1} \\[5mm] & = n\bracks{z^{n - 1}} \pars{1 + z}^{n}\, \bracks{\pars{1 + z}^{n - 1} + \pars{n - 1}z\pars{1 + z}^{n - 2}} \\[5mm] & = n{2n - 1 \choose n - 1} + n\pars{n - 1}{2n - 2 \choose n - 2} \end{align}

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\begin{align} {\cal I}_{3} & = \bbox[5px,#ffd]{\sum_{k = 0}^{n}k^{3}{n \choose k}^{2}} \\[5mm] & = -\,{1 \over 4}\,n^{3}{2n \choose n} + {3 \over 2}\,n \bracks{n{2n - 1 \choose n - 1} + n\pars{n - 1} {2n - 2 \choose n - 2}} \end{align} It can be simplified as \begin{align} {\cal I}_{3} & = \bbox[5px,#ffd]{\sum_{k = 0}^{n}k^{3}{n \choose k}^{2}} = \bbx{{1 \over 2}\,n^{2}\pars{n + 1}{2n - 2 \choose n - 1}} \\ & \end{align}

Answer 3

$$\sum_{j=0}^{n-1} {{n-1}\choose j}x^j = (1+x)^{n-1} \ (*) \\ \sum_{j=0}^{n-1} {{n-1}\choose j}x^{n-1-j} = (1+x)^{n-1} \ (\star)$$ Multiplying these equalities together gives $$x^{n-1} \sum_{j=0}^{n-1} {{n-1}\choose j}^2 + \dots= (1+x)^{2n-2}={{2n-2}\choose{n-1}} x^{n-1} +\dots$$ Comparing coefficients of $x^{n-1}$, you can get the second sum. As for the first sum, start by differentiating (*) and then multiplying by $x$ to get $$\sum_{j=0}^{n-1} j{{n-1}\choose j}x^j =(n-1)x (1+x)^{n-2}$$ Multiplying this with $(\star)$ leads to $$x^{n-1} \sum_{j=0}^{n-1} j{{n-1}\choose j}^2 + \dots= (n-1)x(1+x)^{2n-3} $$ Now you can easily find the coefficient of $x^{n-1}$ in the RHS, compare both sides and you’re done.

Answer 4

You've gotten a lot of feedback on how to evaluate the sum without Vandermonde's convolution $$\sum_{k=0}^r \binom{m}{k} \binom{n}{r-k} = \binom{m+n}{r}$$ but if you want to use it, this is how you would proceed: $$\begin{align} \sum_{j=0}^n \binom{n}{j}^2 &= \sum_{j=0}^n \binom{n}{j} \binom{n}{n-j} = \binom{2n}{n}. \end{align}$$ We simply choose $m = r = n$, but before doing so, we use the binomial coefficient reflection identity $$\binom{n}{j} = \frac{n!}{j!(n-j)!} = \binom{n}{n-j}.$$