Show that
Ʃn i2 = n(n+1)(2n+1)/6, n>=1
i=1
Should I use the formula
Ʃn i = n(n+1)/2
i=1
and if so, how would I go about doing this?
Yes, this works.
Now, use this simple fact: $\displaystyle i(i+1)=\frac{1}{3}\left( i(i+1)(i+2)-i(i-1)(i+1) \right).$
So the sum becomes $\displaystyle \sum\limits_{i=1}^{n}{i(i+1)}=\frac{1}{3}\sum\limi ts_{i=1}^{n}{\left( i(i+1)(i+2)-i(i-1)(i+1) \right)},$ now this is
$\displaystyle \underbrace{\sum\limits_{i=1}^{n}{i^{2}}}_{\text{Y ou need this}}+\underbrace{\sum\limits_{i=1}^{n}{i}}_{\tex t{You know this}}=\frac{1}{3}\underbrace{\sum\limits_{i=1}^{n }{\left( i(i+1)(i+2)-i(i-1)(i+1) \right)}}_{n(n+1)(n+2),\text{ because the above telescopes}}.$
I think you can take it from there.