<s>[INST] 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.. 
 [/INST] 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 $n$.
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