Prove that $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$ I know that it is true but not sure how to write the proof for: $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$. I need help to guide me in the right direction.
Thanks in advance.
edit:
Okay at n=k I have $ 1+4+9 ... + k^2 = (k/6)(k+1)(2k+1)$ 
and at $n=k+1$ I have $((k+1)/6))((k+1)+1)(2(k+1)+1) = ((k+1)/6)(k+2)(2k+3).$
Does my base step need to begin at $n=0$ or $n=1?$ How do I tell? 

When $n=1$, it is Okay.
Suppose that $n=k$, $$1 + 4 + 9 \cdots + k^2 = (k/6)(k+1)(2k+1).$$
Then $n=k+1$,  $$1 + 4 + 9 \cdots + k^2+(k+1)^2 = (k/6)(k+1)(2k+1)+(k+1)^2=((k+1)/6)(k+2)(2k+3).$$
