Using induction to verify a statement I have to prove that this statement is true.
For $n = 1, 2, 3, ...,$ we have $ 1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6$
Basically I thought I'd use induction to prove this. Setting $n = p+1$, I got this so far:
Left hand side: $1² + 2² + 3² +...+ p² + (p+1)² = p² + (p+1)² = 2p² + 2p + 1$
Right hand side: $(p + 1)(p + 1 + 1)(2 * (p + 1) +1)/6$
This is all I've got. I've never worked with induction so I don't even know if this correct to begin. Any insight or solution to my problem is appreciated.
 A: For proving an expression through induction. Check for the base case $n=1$,
Let $P(n)$ is true, $\sum_1^nk^2=1+4+9+ \dots n^2= P(n)=\dfrac{n(n+1)(2n+1)}{6}$. 
You need to prove that $\sum_1^{n+1} (k)^2=1+4+9+ \dots (n+1)^2=P(n+1)=\dfrac{{(n+1)}{}(n+2)(2n+3)}{6}.$     
Hint: $(k+1)^2=k^2+2k+1$ and $\sum_1^n k=\dfrac{n(n+1)}{2}, \sum_1^n 1=n$
A: First of all, your statement
$1^2+2^2+⋯+p^2+(p+1)^2=p^2+(p+1)^2$
is false.  You have simply erased $1^2 + 2^2 + \cdots + (p-1)^2$ from the equation, which you really can't do.
It seems like you are trying to plug in $(p+1)$ to the LHS and the RHS of the equation.  That's fine, but now what you have to do is show that if the LHS and RHS side are equal when $n = p$, they are also equal when $n = p+1$.  So, you've calculated the new LHS and RHS for $n = p+1$ -- now show they're equal.
Inceptio's answer should be a sufficient hint at this point.
A: I think you mean $$\begin{array}{rcl}1^2+2^2+\cdots+p^2+(p+1)^2 &=& \sum_{k=1}^p k^2+(p+1)^2\\ & = & \frac{p(p+1)(2p+1)}{6}+(p+1)^2\end{array}$$
In other words by writing just $1^2+\cdots +p^2+(p+1)^2=p^2+(p+1)^2$, the first $p-1$ terms from the left-hand side are missing, so it's not a valid equality.
At any rate, tidying up the displayed equation above, you should get what you need.
