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- May 9th 2009, 09:06 PM #1

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## Trival question.

How do we show this.

I realized that

This is the same thing as

And the 1's , 3's and 5's cancel out but i'm left with a bunch of 2n's

Here is my reasoning ...

And i assumed that there must be

Leaving me with

Is there a clearer way to show that there are exactly (n/2) number of 2n's .

- May 9th 2009, 10:31 PM #2

- May 9th 2009, 10:43 PM #3

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- May 10th 2009, 06:40 AM #4