User talk:Calogero

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Cheers! Joe (talk) 15:26, 5 August 2015 (UTC)

Theorem
Let $\N$ be the natural numbers.

Then: $\forall n, r \in \N_{> 0}$

$n^r$ can be expressed as the sum of $n$ consecutive odd numbers, the first of which is $n^{r-1} - n + 1$

Proof
$n^r =n*n^{r-1}$

$=\sum_{i=0}^{n-1} n^{r-1}$

$=\sum_{i=0}^{n-1} n^{r-1} - n - 1 + n + 1$

$=\sum_{i=0}^{n-1} n^{r-1} - n + 1 + n - 1$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + \sum_{i=0}^{n-1} n-1$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + n(n-1)$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + 2\frac{n(n-1)}{2}$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + 2\sum_{i=0}^{n-1} i$

$=[\sum_{i=0}^{n-1} n^{r-1} - n + 1] + \sum_{i=0}^{n-1} 2i$

$=\sum_{i=0}^{n-1} n^{r-1} - n + 1 + 2i$

Proof that $n^{r-1} - n + 1$ is always odd
n is either even or odd

if n is even, $n = 2m$, then $(2m)^r$ for any power $r$ is also even

since the difference of two even numbers is even, so $(2m)^r - 2m$ is even

hence, $(2m)^r - 2m + 1$ must be odd

alternately, if n is odd, $n = 2m+1$, then $(2m+1)^r$ is also odd

since the difference of two odd numbers is even, $(2m+1)^r - (2m+1)$ is even

and again, $(2m+1)^r - (2m+1) + 1$ must be odd

Background
For squares, $r=2$, the formula reduces to the sum of the first $n$ odd numbers, a fact which I had known since my youth.

But then I was reading a history of Mathematics, in which I learned that Nichomachus of Gerasa observed that cubical numbers are the sum of successive odd numbers. The example given in the book was $2^3 = 8 = 3+5$

$3^3 = 27 = 7 + 9 + 11$

$4^3 = 64 = 13 + 15 + 17 + 19$

Noticing that the value for $n^3$ was the sum of $n$ consecutive odd numbers, I was curious to see if this worked for $n^4$. It was in playing with fourth powers that I noticed the pattern, generalized it, and then derived the proof.

By Maj Calogero S Cumbo, BSc CD