Is there a close form expression for the series \begin{equation} \sum_{k=0}^n{\alpha \choose k}^2\lambda^k,\quad \alpha ~ \text{is non-integer} \end{equation}
As far as I know, there is an identity involving binomial coefficients \begin{equation} \sum_{k=0}^n{n \choose k}^2 = {2n \choose n} \end{equation} which can be proved using generating function $(1+x)^n(1+x)^n=(1+x)^{2n}$. However similar method can not be applied to the first series.
Is there a way to deal with it ? Thank you !
A standard expression for the $n$th Legendre polynomial is $\frac{1}{2^n}\sum_0^n{n\choose k}^2(x-1)^{n-k}(x+1)^k$.
So that gives immediately the expression $$(1-x)^nP_n\left(\frac{1+x}{1-x}\right)$$