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**╔(σ_σ)╝** $\displaystyle 1 +3 +5 + ... (2n-1) = n^{2}$

How do we show this.

I realized that

This is the same thing as

$\displaystyle 1 +3 +5 + ... + (2n-5) + (2n-3)+(2n-1) = n^{2}$

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 ...

$\displaystyle 1 +3 +5 + ......+ k + (2n-k) ....... + (2n-5) + (2n-3)+(2n-1) = n^{2}$

$\displaystyle k < n$

And i assumed that there must be $\displaystyle \frac{n}{2} 2n's $

Leaving me with $\displaystyle \frac{n}{2} 2n = n^{2} $

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