induction proof: $\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$ I encountered the following induction proof on a practice exam for calculus:
$$\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$$
I have to prove this statement with induction.
Can anyone please help me with this proof?
 A: If $P(n): \sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6},$
we see $P(1): 1^2=1$ and $\frac{1(1+1)(2\cdot1+1)}{6}=1$ so, $P(1)$ is true
Let $P(m)$ is true, $$\sum_{k=1}^mk^2 = \frac{m(m+1)(2m+1)}{6}$$
For $P(m+1),$
$$ \frac{m(m+1)(2m+1)}{6}+(m+1)^2$$
$$=\frac{m(m+1)(2m+1)+6(m+1)^2}6$$
$$=\frac{(m+1)\{m(2m+1)+6(m+1)\}}6$$
$$=\frac{(m+1)(m+2)\{2(m+1)+1\}}6$$  as $m(2m+1)+6(m+1)=2m^2+7m+6=(m+2)(2m+3)$
So, $P(m+1)$ is true if $P(m)$ is true
A: \begin{align}
\sum_{k=1}^{n+1} k^2 & = \left(\sum_{k=1}^n k^2\right) & {} + (n+1)^2 \\[10pt]
& = \underbrace{\left(\frac{n(n+1)(2n+1)}{6}\right)} & {} + (n+1)^2\tag{1}
\end{align}
What you need is the same expression that you see over the $\underbrace{\text{underbrace}}$ but with $n+1$ in place of $n$.  That would be
$$
\frac{[n+1]\Big([n+1]+1\Big)\Big(2[n+1]+1\Big)}{6}.\tag{2}
$$
So the problem is to show that $(1)$ is equal to $(2)$.  If you can be more explicit about where you ran into difficulties, I could possibly say more.
A: HINT:
$$\sum_{k=1}^{n+1}k^2=\sum_{k=1}^nk^2+(n+1)^2=\frac{n(n+1)(2n+1)}{6}+(n+1)^2$$
Now simplify and show that it is equivalent to replacing $n$ by $n+1$ in the original formula.
