Apostol's proof for the method of exhaustion

Question

I'm a high school student currently going through Apostol's calculus. I'm not that familiar with proofs, but I learned Calculus in school up to partial fractions, but we focused more on problems instead of the concept/proof, so please bear with me. I already understand most of the method of exhaustion, but there's one thing that keeps bothering me.

Apostol wants to prove that for every integer n>=1

$$1^2+2^2+...+n^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$$

To do that, he starts with the formula: $$(k+1)^3=k^3+3k^2+3k+1$$

Now, my question is, why did he start with this formula? For instance, if I didn't know anything about this proof and I wanted to start from scratch, what sort of thought process do I go through to come up say "Okay, to prove this idenity we start with this formula."

Answer 1

A thing that suggests that you should be aiming at a cubic polynomial is the fact that $1+2+\cdots+n$ is a quadratic expression on $n$, and now you are just one degree above that.

On the other hand, if indeed you have$$1^2+2^2+\cdots+n^2=P(n)$$for some cubic polynomial $P(x)$ and you want to prove this by induction, it is conveniente to know $P(n+1)-P(n)$, since you know that$$\bigl(1^2+2^2+\cdots+(n+1)^2\bigr)-\bigl(1^2+2^2+\cdots+n^2\bigr)=(n+1)^2.$$So, you want to have$$P(n+1)-P(n)=(n+1)^2.$$

Answer 2

Notice that

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

Lets see what is special about this expression.

1.This expression contains the term $k^2$ 2.We can sum both sides (except ofcourse $\sum_{i=0}^n i^2$

3.We can then simplify the equation to reach the desired result.

Note: We could have used $(k+2)^3-k^3$ or for finding $\sum_{k=1}^n k^4$ we can sum both sides of $\frac{(k+1)^4+(k-1)^4}{2}=k^4+6k^2+1$ . You can use this technique for finding many different summitions.