I am looking to calculate the following sum: $$ \sum_{k=0}^N{N\choose k}^2x^k $$ The sum appears, e.g., in the context of calculating the partition function of a DNA, where the nucleotides in the two parallel strands can form $k$ pairs with pairing probability $p$ and energy $E_k=-k\Delta$ (assuming that the bonds do not cross). Then the partition function (average over nucleotide configurations) is given by $$ Z(\beta) = \sum_{k=0}^N\sum_{k=0}^N{N\choose k}^2p^ke^{\beta k \Delta} $$
I know that $$ \sum_{k=0}^N{N\choose k}^2 = {2N\choose N}, $$ and it seems that there should be a simple way to get to the desired identity... on the other hand, the simplicity might be misleading.
$$S_n(x)=\sum_{k=0}^n{n\choose k}^2x^k=(1-x)^n \,P_n\left(\frac{1+x}{1-x}\right)$$ where appear Legendre polynomials.