# 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 $k^2$:

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

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

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

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

• $$\sum_{k=1}^n(k + 1)^3 - \sum_{k=1}^nk^3=\sum_{j=2}^{n+1}j^3 - \sum_{k=1}^nk^3=(2^2+3^3+\cdots+n^3+(n+1)^3)-(1^3+2^3+\cdots+(n-1)^3+n^3)=(n+1)^3-1^3$$ – Did Dec 16 '16 at 19:36
• You can also find the formula for the sum of squares this by writing $i^2 = 2( \sum_{k=1}^i k) - i$ and interchanging summation. I tend to think this is less opaque of a method. You can also repeat it to inductively find the sum of (m + 1)th powers knowing the formula for the sum of the $j$th powers for all $j \le m$, although the algebra becomes more complicated as you go along. – Vik78 Dec 16 '16 at 20:59
• If you just want a derivation of the formula, this question has been asked and answered many times, for example math.stackexchange.com/questions/183316/… – David K Dec 17 '16 at 2:26

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.

• Great! Makes sense. I should remember to expand the summation notation when attempting to solve problems like this one. – Joseph Dasenbrock Dec 16 '16 at 22:10

HINT:

For a Telescoping Series we have

$$\sum_{k=1}^n (a_{k+1}-a_k)=a_{n+1}-a_1$$

Now, set $a_k=k^3$.

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*}

• that was a helpful way of looking at it, solving from the end instead of the beginning. – Joseph Dasenbrock Dec 17 '16 at 0:10
• You can continue this bootstrap method and derive the sum of any powers you have the patience to compute. – Laars Helenius Dec 17 '16 at 1:14