using theorem k (m) (n) = (m + n) r=0 (r) (k-r) (k) [ please...

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using theorem k Σ (m) (n) = (m + n) r=0 (r) (k-r) (k) [ please consider (n) as combinations] (r) n 1.16) show that Σ (n)² = (2n) r = 0 (r)² (n) using theorem k Σ (m) (n) = (m + n) r=0 (r) (k-r) (k) [ please consider (n) as combinations] (r) n 1.16) show that Σ (n)² = (2n) r = 0 (r)² (n)

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