Math Number Sequence

[Makalo]

I am learning or have learned a bit about squaring integers and discovered a pattern of odds numbers that increase by 2.
I'll give an example

1 x 1 = 1, 1 + 3 = 4 or 2 square
2 x 2 = 4, 4 + 5 = 9 or 3 square
3 x 3 = 9, 9 + 7 = 16 or 4 square
4 x 4 = 16, 16 + 9 = 25 or 5 square
5 x 5 = 25, 25 + 11 = 36 or 6 square

The interesting thing about this is any sequence i use such as 3, 6, 9, 12, 15

3 x 3 = 9
    difference between 9 & 36 is 27, ---- 9 x 3 = 27,
6 x 6 = 36
    difference between 36 & 81 is 45, ---- 9 x 5 = 45,
9 x 9 = 81
    difference between 81 and 144 is 63, ---- 9 x 7 = 63,
12 x 12 = 144
    difference between 144 and next number is 81, ---- 9 x 9 = 81,

Now if I multiply the 27 x 3 i get the value of the square of the 3rd number in series which is 81 and the square root of 81 is 9.

This "little" discovery works with just about any sequence of numbers from 1-100 of 1 count to multipliers of a specific number or even the odd number sequence.

There is also another thing which I found a little strange, its like a times table chart.

     0   1   2
1   1   6   11
3   2   7   12
5   3   8   13
7   4   9   14
9   5  10  15

I thought I might be able to do something with the odd number sequence like I figured out how to square just about any number with a 5 in the ones place.

12345 - 5 = 12340
12340 /  5 = 2468
2468 / 2 = 1234
1234 x 1235 = 1523990
Then append 25 to the answer
12,345 squared is 152,399,025

Understandably we could have just multiplied 12345 x 12345 and gotten the answer, but when you write code you sometimes want to manipulate the process of a calculation.

So I am curious if there is any special meaning mathematical or otherwise around this sequence of odd numbers?

 

[Lordfire]

1 x 1 = 1, 1 + 3 = 4 or 2 square
2 x 2 = 4, 4 + 5 = 9 or 3 square
3 x 3 = 9, 9 + 7 = 16 or 4 square
4 x 4 = 16, 16 + 9 = 25 or 5 square
5 x 5 = 25, 25 + 11 = 36 or 6 square

Please mind you are writing:

n*n + n*2 + 1 == n*(2 + n) + 1 == n^2 + 2n + 1

Which is the expanded form of:

(a + b)^2 == a^2 + 2ab + b^2

Substituting terms

(n + 1)^2 = n^2 + 2n + 1

You didn't discover anything new, it's just calculus algebra (thanks @Damc)

 

[hodleo]

Nobody cares anyway.