Compute (1^2 + 2^2 + 3^2 + ... + 10^2)/770.
sumfor(n, 1, 10, (n^2))=462 / 770 =3 / 5
In general
\(\Sigma_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\)
When n = 10 we have:
\(\Sigma_{k=1}^{10}k^2=\frac{10.11.21}{6}=385\)
And 385/770 = 1/2
