TITLE: Evaluation of $\sum_{k=0}^n{n\choose k}^2u^k$
QUESTION [7 upvotes]: I am trying to evaluate the finite sum
\begin{equation}
f(u)=\sum_{k=0}^n{n\choose k}^2u^k,\quad 0<u\le1
\end{equation}
In an first attempt, I think of the generating function
\begin{equation}
(1+x)^n(u+x)^n = \sum_{k\ge0}{n\choose k}x^k\sum_{k\ge0}{n\choose k}u^kx^{n-k}=\cdots+x^n\sum_{k\ge0}{n\choose k}^2u^k+\cdots
\end{equation}
which means that $f(u)$ is the coefficients of the term $x^n$.
Expanding $(1+x)^n(u+x)^n$ into $[1+(1+u)x+x^2]^n$, and using the multinomial expansion, I get an expression for the coefficient of $x^n$. However, such an expression is more complicated than $f(u)$. It seems I am making the problem even more difficult.
Can someone help me find the value of $f(u)$.
Thank you.

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
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$\ds{\fermi\pars{u}\equiv\sum_{k = 0}^{n}{n \choose k}^{2}u^{k}:\ {\large ?}.
     \qquad 0\ <\ u\ \leq\ 1}$.

\begin{align}
\fermi\pars{u}&=\sum_{k = 0}^{n}{n \choose k}u^{k}{n \choose n - k}
=\sum_{k = 0}^{n}{n \choose k}u^{k}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{n - k + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{n + 1}}
\sum_{k = 0}^{n}{n \choose k}\pars{zu}^{k}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{n} \over z^{n + 1}}
\,\pars{1 + zu}^{n}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}
{\bracks{uz^{2} + \pars{u + 1}z + 1}^{n} \over z^{n + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{a,b,c=0 \atop a + b + c=n}{n \choose a,b,c}u^{a}\pars{u + 1}^{b}
\oint_{}{1 \over z^{n - 2a - b + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\left.%
\sum_{a,b,c=0 \atop a + b + c=n}{n \choose a,b,c}u^{a}\pars{u + 1}^{b}\right\vert
_{\,n\ -\ 2a\ -\ b\ =\ 0}
=\sum_{c = 0}^{\floor{n/2}\atop}{n! \over \pars{n - 2c}!\,\pars{c!}^{2}}u^{c}\pars{u + 1}^{n - 2c}
\end{align}

$$\color{#66f}{\large%
\fermi\pars{u}=
n!\,\pars{1 + u}^{n}\sum_{c = 0}^{\floor{n/2}\atop}{1 \over \pars{n - 2c}!\,\pars{c!}^{2}}\bracks{u \over \pars{1 + u}^{2}}^{c}}
$$