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Prove that sum(r=0)^(2n) r.(""^(2n)C(r))...

Prove that `sum_(r=0)^(2n) r.(""^(2n)C_(r))^(2)= 2.""^(4n-1)C_(2n-1)`.

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To prove that \[ \sum_{r=0}^{2n} r \cdot \binom{2n}{r}^2 = 2 \cdot \binom{4n-1}{2n-1}, \] we will follow these steps: ...
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Prove that sum_(r=0)^(2n)(r. ^(2n)C_r)^2=n^(4n)C_(2n) .

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Prove that sum_(r = 0)^n r^2 . C_r = n (n +1).2^(n-2)

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