Observe that all natural numbers n^3 can be written as a sum of n odd integers that are revolving around n^2 For example,
$1^3$ = $1$,
$2^3$ = $(2^2-1) + (2^2+1)$ 
$3^3$ = $(3^2-2) + (3^2) + (3^2 + 2)$ 
$4^3$ = $(4^2-3) + (4^2 - 1) + (4^2 + 1) + (4^2 + 3)$ 
I have been able to deduce a general formula for this behavior, which would be:
$n^3 = (n^2 - (n-1)) + (n^2 - (n-3)) + ... + (n^2 + (n-3)) + (n^2 + (n-1))$
I am then asked to prove this general formula via induction and I believe applying the second principle of induction would be more prudent than the first.
I read an explanation about it here: What is the second principle of finite induction? 
It tells me that I have to assume all $n$ from 1 to $k$ is true instead of just $n = k$, so I assume I have to factor multiple instances of $i$ (where $i$ is an integer from 1 to $k$) to prove this general formula, but I don't really know how to approach it.. 
 A: After cancelling what cancels, you just have $n^3=n^2\cdot n.$
(hopefully helpful) edit:
We can apply arithmetic sequence results. If an arithmetic sequence has first term $a,$ common difference $d,$ last term $l$ and number of terms $n,$ then for $1 \le k \le n$ the $k$th term $t_k$ is $a+(k-1)d.$ [so last term $a+(n-1)d.$]
The sum is then $(1/2)(a+l)n.$ All these can be shown by induction, in fact I don't see how otherwise to use induction directly for your problem since so much cancels.
Anyway apply these to your sequence with $a=n^2-(n-1)$ and $d=2,$ with $n$ terms. Then $a$ is odd since $n^2$ and $(n-1)$ have opposite parity, then since $d=2$ all terms are odd. Last term is now $[n^2-(n-1)]+2(n-1)=n^2+(n-1),$ so that $(1/2)(a+l)=n^2$ giving sum $s=n^2\cdot n=n^3.$
A: Note that:
In the expression
$n^3 = (n^2 - (n-1)) + (n^2 - (n-3)) + ... + (n^2 + (n-3)) + (n^2 + (n-1))$,
$n-1$ cancels from the first term and the last term.
$n-3$ cancels from the second term and the second to last term.
$n-5$ cancels from the thirst term and the third to last term.
And so on.
$n-a$ cancels from the $(\frac{a+1}{2})^\text{th}$ term and the $(\frac{a+1}{2})^\text{th}$ to last term, for any odd positive integer $a$.
We left with $n^3=n^2+n^2+n^2+\dots$ $[n \text{ times}]$. This means that $n^3=n\cdot n^2$, which is true for any $n$.
