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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.

Roger V.
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  • Possible repeat of this post: https://math.stackexchange.com/questions/148583/combinatorial-proof-of-summation-of-sum-limits-k-0n-n-choose-k2-2n – klein4 May 26 '21 at 17:02
  • @klein4 here I am asking for a more general result. – Roger V. May 26 '21 at 17:09
  • Fun question +1. some more background would be nice. Although bio is not the main thing here , so this is rather a personal question I guess. maybe a link in the comment is better and easier – mick Feb 16 '23 at 22:58

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$$S_n(x)=\sum_{k=0}^n{n\choose k}^2x^k=(1-x)^n \,P_n\left(\frac{1+x}{1-x}\right)$$ where appear Legendre polynomials.

Claude Leibovici
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