# Summation formula

• February 27th 2008, 11:44 PM
HallsFB
Summation formula
Hello

Prove that the sum of k^2 from k=1 to n is equal to:

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

The book provided a hint: k^3 -(k -1)^3 = 3k^2 -3k + 1.
But I don't know how to use it.

$(v+1)^{3} - v^{3} = 3v^{2} + 3v+1$
Write this equation down for $v = 0..n$ and add them.
Then $(n+1)^{3} = 3S_2 + 3S_1+n+1$ and $3S_{2} = (n+1) \left[(n+1)^{2} - 1 -\frac{3}{2}n \right] = \left(n+1\right)\left(n^{2} + \frac{1}{2}n \right)$
$S_2 = \frac{1}{6}n(n+1)(2n+1)$