Has anyone seen this?

Question

I noticed that whole numbers squared increase by the odd numbers. Halves increase by even numbers. Has anyone seen this in books anywhere?
1^2=1
2^2=4 . . . 4-1=3
3^2=9 . . . 9-4=5
4^2=16. . . 16-9=7

3, 5, 7, 9, 11, 13, 15

This pattern continues forever.

Also 0.5 squared equals 0.25,
1.5 squared equals 2.25,
2.5 squared equals 6.25
3.5 squared equals 12.25
2.25-0.25=2
6.25-2.25=4
12.25-6.25=6

2, 4, 6, 8, this pattern also continues.

I apologize for any mistakes, but I'm too tired to proofread this.

Answer 1

You are absolutely correct.

In fact I'll prove as to why that is:

(All I'm doing is doing what you did but 'vaguely', you'll see)

Let's say x is a whole number, so x+1 is the next one, Now let's square both:

x^2 = x^2
(x+1)^2 = x^2 + 2x + 1

If you subtract them you get:

2x+1 >>> which is a formula for odd numbers


Now let's do the same for the halves:

((x+1)/2)^2 = (x^2+2x+1)/4
((x+1)/2 + 1)^2 = (x^2+6x+9)/4

Again let's subtract them:

4x+8 >>> An expression that assures an even number

There you go, mathematical proof instigated from interesting patterns

Answer 2

You are looking for a pattern in (b^2 - a^2), where b = a+1
b^2 - a^2 = (b+a)(b-1) = (2a+1)(1) = 2a+1 = even number + 1, which by definition is an odd number when a is an integer.

What if a = integer + 0.5? Then 2a + 1 = 2 (integer + 0.5) + 1
= 2(integer) + 1 + 1 = 2(integer + 1) = even number, by definition.

Answer 3

Yes, the pattern works. It can be proved algebrically. If n * n = n^2, the next larger integer is n+1. (n+1)^2= n^2+2n+1. So in terms of n, the difference is 2n+1.

Answer 4

Sure. The relation is easily proved by induction. Use (n+1)^2 = n*2 + 2n + 1, noting that 2n+1 is an odd number.

Answer 5

Wow...you are right!

I never saw this pattern, and I look for patterns often!

Good find.