Finding out the square of any number (Imhappy62789's technique)

Heres a list of squares, this is a bit important I promise.

$1^2$ $2^2$ $3^2$ $4^2$ $5^2$ $6^2$ $7^2$ ...

...and so on.

Heres the story behind it, one day in school there was a sheet of all the squares, and I thought something. To get from $1^2$ to $2^2$ , think of it like a staircase. $1^2$ is $1\cdot1$ and then from that you go to $1\cdot2$ , which you add another $1$ to it, and then $2\cdot2$ , which you add a $2$ , since you're increasing the factor. So adding those two numbers we said earlier to $1^2$ , we get $1^2 + 1 + 2$ , which equals $4$ , the square of $2$ . About 6 months later, I thought of something else. What if we can get from $1^2$ to, say, $3^2$ using the same method? So I did the same thing. $1^2$ is equal to $1\cdot1$ , then go up to $1\cdot2$ , to add $1$ to it, then up to $2\cdot2$ , to add a $2$ to it, and then to $2\cdot3$ , which adds another $2$ , then to $3\cdot3$ , to add a $3$ to it. So it gives us this: $1^2 + 1 + 2 + 2 + 3$ , which simplified gives us this: $1 + 1 + 2 + 2 + 3$ . Now, I noticed something here, it looks just like an arithmetic sequence, or series, that is. So I thought, we can find the square of any integer, if we develop a formula for it. So first I made some variables. Let $x =$ the number of the number that is getting squared. Let $n$ be the number of numbers in the arithmetic sequence, or series. So $n = x - 2$

So, setting up the arithmetic sequence, or series, looks like this:

$1 + 2 + 2 + 3 + 3 + 4 + 4 + \cdots + x - 1 + x - 1 + x$

So, I made a formula, first we start with $1^2$ , and then adding $1$ to it (so ignoring the $1$ and pretending it's not there in the arithmetic sequence or series), and then adding the arithmetic series formula, which is $\left(2 + \left(x - 1\right)\right) \cdot \frac{n}{2}$ Notice that its basically two arithmetic sequences, or series, so you multiply it by $2$ , which makes

$\left(2 + \left(x - 1\right)\right) \cdot \frac{n}{2} \cdot 2$ , which simplified gives us

$\left(2 + \left(x - 1\right)\right) \cdot n$

But, we also have to add the last one, which is $x$ to it, so our formula turns into this

$1^2 + 1 + \left(2 + \left(x - 1\right)\right) \cdot n + x$

Simplified, it turns into this

$1^2 + 1 + \left(2 + x - 1\right) \cdot n + x$ $=$ $1^2 + 1 + \left(2 - 1 + x\right) \cdot n + x$

So our completed formula would be:

$1^2 + 1 + \left(1 + x\right) \cdot n + x$

If you have a -2 as the exponent instead, just put a 1 over everything, so it should be like this, $\frac{1}{1^2 + 1 + \left(1 + x\right) \cdot n + x}$

This is basically my first mathematical discovery, and I have absolutely no idea if anyone has made it before, if you know if this has been posted before, please put the paper or anything in here to link towards it.

This article has been proposed for deletion. The reason given is: lacks notability, long title (move to user namespace). Sysops: Before deleting this article, please check the article discussion pages and history.

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Finding out the square of any number (Imhappy62789's technique)