Given $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, find a compact formula for $\sum_{i=1}^n i^2$ My first post here, I read the introduction materials, hopefully this is a good question. 
I'm working through a textbook on proofs (Chartrand) in preparation for taking some higher-level mathematics courses next semester. This is pretty much my first exposure to proofs, and I'm refreshing myself on a lot of basic math concepts. I've taken up to Calc II at a university, but it has been a while. 
I know there are a lot of identities which I need to brush up on. I think I'm missing some identities for summations, but I can't figure out what they are. 
Here is a problem that I'm having trouble understanding:

Given $\sum_{i=1}^n i = \frac{n(n+1)}{2}$, find a compact formula for
  $\sum_{i=1}^n i^2$

The example given in Chartrand starts here: 

$(k + 1) ^3 = k^3 + 3k^2 + 3k +1$

Solve for $k^2$:

$k^2 = \frac{1}{3}\left[(k + 1)^3 - k^3\right] - k - \frac{1}{3}$

Add summation to both sides: 

$\sum_{i=1}^n k^2 = \frac{1}{3}\left[\sum_{i=1}^n(k + 1)^3 - \sum_{i=1}^nk^3\right] - \sum_{i=1}^nk - \frac{1}{3}\sum_{i=1}^n 1$

I understand how to get to the above point, the below transition is where I'm stuck:

$k^2 = \frac{1}{3}\left[(n + 1)^3 - 1^3\right] - \frac{1}{2} n(n + 1) - \frac{1}{3}n$

The main thing I don't understand is how to go from 

$\sum_{i=1}^n(k + 1)^3 - \sum_{i=1}^nk^3$

to

$(n + 1)^3 - 1^3$

Thanks in advance for input.
 A: What I don't understand is how to go from $\sum\limits_{i=1}^n(k+1)^3-\sum\limits_{i=1}^nk^3$ to $(n+1)^3-1^3$
Using two lines to hopefully make it more obvious, we have:
$\sum\limits_{i=1}^n(k+1)^3-\sum\limits_{i=1}^nk^3 =$
$\begin{array}{}&2^3&+3^3&+4^3&+\dots&+n^3&+(n+1)^3\\-(1^3&+2^3&+3^3&+4^3&+\dots&+n^3)\end{array}$
Each of the terms $2^3,3^3,4^3,\dots$ on up to $n^3$ cancel leaving only the $(n+1)^3$ and the $1^3$ remaining.
A: HINT:
For a Telescoping Series we have
$$\sum_{k=1}^n (a_{k+1}-a_k)=a_{n+1}-a_1$$
Now, set $a_k=k^3$.
A: Here is another way to organize the work that might give some insight:
$$
\begin{align*}
1+\sum_{k=1}^n(k+1)^3&=\sum_{k=1}^nk^3 + (n+1)^3\\
1+\sum_{k=1}^n(k^3+3k^2+3k+1)&=\sum_{k=1}^nk^3 + (n+1)^3\\
1+\sum_{k=1}^nk^3+3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+\sum_{k=1}^n1&=\sum_{k=1}^nk^3 + (n+1)^3\\
3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+\sum_{k=1}^n1&=(n+1)^3-1\\
3\sum_{k=1}^nk^2+\frac{3n(n+1)}{2}+n&=(n+1)^3-1\\
\sum_{k=1}^nk^2&=\frac{2(n+1)^3-3n(n+1)-2n-2}{6}\\
\sum_{k=1}^nk^2&=\frac{n(n+1)(2n+1)}{6}.\\
\end{align*}
$$
