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