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Alternating binomial-square sum

Coefficient extraction from

(1 - x)^{n} (1 + x)^{n} = (1 - x^{2})^{n}

gives

\sum_{r = 0}^{n} (- 1)^{r} (r n)^{2} = {0, (- 1)^{n /2} (n /2 n), n odd, n even .

 Ancestors (6)