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I got stuck at the computation of the sum . the problem is K . how to simply this combinatoric ? Is this problem sorted out by directly substituting the k value? In general, the number of picks (k) changes, but in this case it is constant.(n) What does this mean? $$

lovmathh
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As OP has now edited to present some of their ideas, here is a very simple proof. We find

$$\sum_{k=0}^n k {2n-k\choose n} = \sum_{k=0}^n k {2n-k\choose n-k} = [z^n] (1+z)^{2n} \sum_{k=0}^n k z^k (1+z)^{-k}.$$

Here the coefficient extractor enforces the upper range of the sum and we have

$$[z^n] (1+z)^{2n} \sum_{k\ge 0} k z^k (1+z)^{-k} = [z^n] (1+z)^{2n} \frac{z/(1+z)}{(1-z/(1+z))^2} \\ = [z^n] (1+t)^{2n} z (1+z) = [z^{n-1}] (1+z)^{2n+1} = {2n+1\choose n-1}.$$

Marko Riedel
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