Finding All Odd Numbers in the Series

[Hyperreality]

Here is a question I suddnely thought of.

The natural number series: 1,2,3...

If we double the series, we get: 2,4,6...

In other words get all the even numbers.

My question is there any ways to determine all odd number series?

 

[Doctor Luz]

If you add one more step and substract 1 after multpliying by 2 to the series you obtain the odd number series.

 

[selfAdjoint]

Or add 1 and treat 1 as a special case. So every even number is of the form 2n and every odd number is of the form 2n+1. Where n can be read as "something".

Now look what you can do. (2n)^2 = 4n^2 = 2(2n^2) so the square of an even number is of the form 2 times something, and so it is even too.

And (2n+1)^2 = 4n^2 + 4n +1
= 2(2n^2+2n) + 1.
So the square of an odd number is one more than an even number, so it is odd.

You have just proved a theorem: The square of an even number is even and the square of an odd number is odd.

 

[quartodeciman]

Take this another step.

Pretend that you are IBM Corporation{1} in the mid-1950s. Start your sequence with 0 instead of 1.

0 1 2 3 4 5 ...

Now, square the members of this sequence.

0 1 4 9 16 25 ...

This is the derived sequence of perfect squares.

Now, subtract adjacent members of this sequence.

Voilá!

1 3 5 7 9 ...

---

{1}IBM computer systems numbered peripherals devices and their plug ports beginning with 0 instead of with 1 (which is what other computer manufacturers started with at that time).

 

[quartodeciman]

One more bit.

Start with

0 1 2 3 4 5 ...

This time, instead of squaring each member, multiply adjacent neighbors.

0 2 6 12 20 30 ...

Now, subtract adjacent terms of this.

2 4 6 8 10 ...

whoopie!

 

[Integral]

Originally posted by quartodeciman
One more bit.

Start with

0 1 2 3 4 5 ...

This time, instead of squaring each member, multiply adjacent neighbors.

0 2 6 12 20 30 ...

Now, subtract adjacent terms of this.

2 4 6 8 10 ...

whoopie!

n(n+1) - n(n-1)=n²+n - n²+n =2n