Evaluation of $\sum_{k=0}^n{n\choose k}^2u^k$ 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.
 A: The Legendre polynomial $P_n$ happens to satisfy
$$P_n(x) = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}^2 (x - 1)^{k}(x + 1)^{n-k}.$$
Hence
$$(1-x)^n P_n\left(\frac{x-1}{x+1}\right) = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k}^2 (2x)^{k} 2^{n-k} =\sum_{k=0}^n \binom{n}{k}^2 x^k.$$
To prove the earlier equation, note that Rodrigues' formula gives
$$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left((x^2 - 1)^n\right).$$
Hence
\begin{align*}
2^n P_n(x) 
  &= \frac{1}{n!} \sum_{k=0}^n \binom{n}{k} \left[\frac{d^k}{dx^k} (x - 1)^n\right] \left[\frac{d^{n-k}}{dx^{n-k}} (x + 1)^n\right] \\
  &= \frac{1}{n!} \sum_{k=0}^n \binom{n}{k}  \left[k! \binom{n}{k} (x - 1)^{n-k}\right] \left[(n-k)! \binom{n}{n-k} (x + 1)^k \right] \\
  &= \sum_{k=0}^n \binom{n}{k}^2 (x - 1)^{n-k} (x + 1)^k \\
  &= \sum_{k=0}^n \binom{n}{k}^2 (x - 1)^{k} (x + 1)^{n-k}.
\end{align*}
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\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}

$$\color{#66f}{\large%
\fermi\pars{u}=
n!\,\pars{1 + u}^{n}\sum_{c = 0}^{\floor{n/2}\atop}{1 \over \pars{n - 2c}!\,\pars{c!}^{2}}\bracks{u \over \pars{1 + u}^{2}}^{c}}
$$
A: The $n$-th Legendre polynomial has the following representation as a finite series (ref. Wolfram functions site):
$$P_n{(z)}=\left(\frac{z-1}{2}\right)^n\sum_{k=0}^{n}\binom{n}{k}^2\left(\frac{z+1}{z-1}\right)^k.$$
Substituting $u=\frac{z+1}{z-1}=1+\frac{2}{z-1}$ into the above formula, yields:
$$P_n{\left(\frac{u+1}{u-1}\right)}=\frac{1}{(u-1)^n}\sum_{k=0}^{n}\binom{n}{k}^2u^k.$$
