sommatorie

... segue subito che

3 S_k:1->n k^2 + 3 S_k:1->n k + S_k:1->n 1 = (n + 1)^3

che significa

3 S2 + 3 S1 + n + 1 = n^3 + 3n^2 + 3n + 1

3 S2 = n^3 + 3n^2 + 3n + 1 - n - 3 n(n + 1)/2 - 1

S2 = n^3/3 + n^2 + n - (n^2 + n)/2 - n/3

S2 = 2n^3/6 + (n^2 + n)/2 - 2n/6 =

= (2n^3 + 3n^2 + 3n - 2n)/6 =

= n (2n^2 + 3n + 1)/6 =

= n(2n^2 + 2n + n + 1)/6 =

= n(2(n+1) + n+1)/6 =

= n(n+1)(2n+1)/6