Evaluation of $\sum_{k=0}^n{n\choose k}^2u^k$

Question

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.

Answer 1

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*}

Answer 2

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:

$$P_n{\left(\frac{u+1}{u-1}\right)}=\frac{1}{(u-1)^n}\sum_{k=0}^{n}\binom{n}{k}^2u^k.$$

Answer 3

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

\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}

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