Observe that all natural numbers n^3 can be written as a sum of n odd integers that are revolving around n^2

Question

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..

Do you have to prove this via induction? It seems that proving directly i.e., showing that the prescribed sum for general $n$ simplifies to $n^3$, seems to be the easier and the more natural way to do this. — Mike, Sep 09 '19 at 17:33
Some call the technique you’re calling second principle strong induction — J. W. Tanner, Sep 09 '19 at 17:38
Yes, the problem specifically states I HAVE to use induction, not via cancellation. My apologies, I should have clarified. — Crystal McMillian, Sep 09 '19 at 18:49
Damn this looks like a neat exercise to use carry-overs base n — Francisco José Letterio, Oct 11 '19 at 04:56

coffeemath · Answer 1 · 2019-09-09T21:01:24.433

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 $s=n^2\cdot n=n^3.$

score 1 · Answer 2 · answered Sep 09 '19 at 17:46

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 $n$.

Observe that all natural numbers n^3 can be written as a sum of n odd integers that are revolving around n^2

2 Answers2