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I need to find the sum $$\sum_{k=0}^{n} \binom{n}{k}^2kp^k,$$ for an integer $n$ and $0<p<1$.

Mathematica would only return $$\sum_{k=0}^{n} k p^k \binom{n}{k}^2 = n^2 p \ _2F_1(1-n, 1-n, 2, p).$$

However, I would prefer a solution in terms of Legendre or Krawtchouk polinomials rather then a hypergeometric function.

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This one-year old answer was downvoted today. I ask the downvoter to explain what exactly makes this answer not useful.

The hypergeometric function you obtain can indeed be expressed in terms of Legendre polynomials.

  1. The differentiation formula $$\frac{d}{dp}{}_2F_1(a,b,c,p)=\frac{ab}{c}{}_2F_1(a+1,b+1,c+1,p)$$ allows to write $$_2F_1(1-n,1-n,2,p)=\frac{1}{n^2}\frac{d}{dp}{}_2F_1(-n,-n,1,p).\tag{1}$$

  2. On the other hand, one has Murphy's formula for Legendre polynomials: \begin{align}P_n(z)&={}_2F_1\left(-n,n+1,1,\frac{1-z}{2}\right)=\\ &=\left(\frac{z+1}{2}\right)^n{}_2F_1\left(-n,-n,1,\frac{z-1}{z+1}\right).\tag{2} \end{align}

  3. Now, combining (1) and (2), we get $$_2F_1(1-n,1-n,2,p)=\frac{1}{n^2}\frac{d}{dp}\left[\left(1-p\right)^nP_n\left(\frac{1+p}{1-p}\right)\right],$$ so that finally $$\boxed{\displaystyle\sum_{k=0}^n k p^k{n\choose k}^2=p\frac{d}{dp}\left[\left(1-p\right)^nP_n\left(\frac{1+p}{1-p}\right)\right]}\tag{3}$$

P.S. If you don't like the derivative in (3), you can express it in terms of $P_n$ and $P_{n-1}$ (or $P_{n+1}$) using the known differentiation formulas, e.g. the formula (44) here.

P.P.S. The formula (3) is an easy consequence of a simpler relation (known to Mathematica): $$\sum_{k=0}^n p^k{n\choose k}^2=\left(1-p\right)^nP_n\left(\frac{1+p}{1-p}\right). $$

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I hate downvoting for nothing! So I vote you up. Last week I had several downvotes for no reason... Voteup for you O.L.!!! – johannesvalks Jul 19 '14 at 9:28

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