I am looking for a closed form for $\displaystyle S= \sum_{k=0}^n x^k \binom{n}{k}^2$. Does there exist such closed form?
Closed form for $\displaystyle S= \sum_{k=0}^n x^k \binom{n}{k}^2$
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$\begingroup$
combinatorics
binomial-coefficients
binomial-theorem
1 Answers
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$$\sum_{k=0}^n x^k \binom{n}{k}^2=(x-1)^n \text{P}_n\left(\frac{x+1}{x-1} \right) $$ $\text{P}_n$ is a Legendre polynomial :
http://mathworld.wolfram.com/LegendrePolynomial.html
This is related to a form of series definition of Legendre polynomials : $$\text{P}_n(z)=\left(\frac{z-1}{2}\right)^n \sum_{k=0}^n \left(\frac{z+1}{z-1} \right)^k \binom{n}{k}^2$$
With $z=\frac{x+1}{x-1}$ leads to the above closed form.