TITLE: Is there anything special about this summation I found that calculates out the square, cube, fourth, etc. power of any integer?
QUESTION [3 upvotes]: let me start by saying that my formatting may be way off, but it's the best I can do, and has little to do with the question, and I will make sure I am as clear as humanly possible, including showing all my work 
So I was playing with squares and stuff, I found that I could calculate out every successive perfect square with: $$\sum_{n=0}^{x-1} (2n+1) = x^2$$ where $$ 0 \le n\le(x-1)|_{||}| 1\le(n+1)\le x$$ 
For example: $5^2=\sum_{n=0}^4(2n+1)=1+3+5+7+9=25$
and then found through trial and error that I could also calculate out successive cubes with: $$ x^3=\sum_{n=0}^{x-1}(3n^2+3n+1) $$ with the same rules for n.
For example: $5^3=\sum_{n=0}^4 (3n^2+3n+1)=1+7+19+37+61=125$
Then I discovered that the cubic formula can be factored so that: $$x^3=\biggr(\sum_{n=0}^{x-1}\bigr((2n+1)*(n+1)\big)\biggr) $$
I noticed each even power can be represented by: $x^{m_{even}}=\biggr(\sum_{n=0}^{x-1}(2n+1)\biggr)^{\frac m 2}$ 
While odd powers can be represented with: $x^{m_{odd}}=\biggr(\sum_{n=0}^{x-1}(2n+1)\biggr)^{\frac{m-1} 2}*(n+1)$ 
where $(n+1)=x$ , but only the final term is replaced with $x$. 
Which eventually lead me to an interesting relation (at least to me): $$x^m=\bigg(\sum_{n=0}^{x-1}(2n+1)\bigg)^{\frac {m-3} 2}*\bigg(\sum_{n=0}^{x-1}(3n^2+3n+1)\bigg)$$
I get that I'm essentially just multiplying squares and cubes together and essentially saying that: $$x^m=\big((x^2)^{\frac {m-3} 2}\big)*x^3$$ and all the math works out to give you: $$x^{m-3}=x^{m-3}$$
However, I want to know if this is something unique, or are there many other ways to represent $x^m$ than what I've found? Is there anything useful somebody could do with something like this, or is it only good for looking cool? 
p.s. Sorry if what I tried to convey didn't make any sense, I'm not familiar with how I'm supposed to phrase things in math.

REPLY [1 votes]: You are implicitly using telescoping series with your sums giving the integer squares and cubes. For the first case, you have
$$\begin{equation}\begin{aligned}
& \sum_{n=0}^{x-1}(2n+1) \\
& = \sum_{n=0}^{x-1}((n^2 + 2n+1) - n^2) \\
& = \sum_{n=0}^{x-1}((n+1)^2 - n^2) \\
& = (1^2 - 0^2) + (2^2 - 1^2) + \ldots + ((x-1)^2 - (x-2)^2) + (x^2 - (x-1)^2) \\
& = x^2 - 0^2 \\
& = x^2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The final result occurs because all of the terms cancel each other out, except for the $x^2$ and $0$ terms. Similarly, you also have
$$\begin{equation}\begin{aligned}
& \sum_{n=0}^{x-1}(3n^2+3n+1) \\
& = \sum_{n=0}^{x-1}((n^3 + 3n^2+3n+1) - n^3) \\
& = \sum_{n=0}^{x-1}((n+1)^3 - n^3) \\
& = (1^3 - 0^3) + (2^3 - 1^3) + \ldots + ((x-1)^3 - (x-2)^3) + (x^3 - (x-1)^3) \\
& = x^3 - 0^3 \\
& = x^3
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Regarding your statement of
$$x^m=\bigg(\sum_{n=0}^{x-1}(2n+1)\bigg)^{\frac {m-3} 2}*\bigg(\sum_{n=0}^{x-1}(3n^2+3n+1)\bigg) \tag{3}\label{eq3A}$$
It's true, as you determined by substituting and simplifying. As for any particular use for this, I don't know of any offhand. Also, regarding other ways to express $x^m$, there are many possibilities, with the particular choice often depending on what you're checking on or trying to determine.
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