1. ## Trival question.

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

2. Originally Posted by ╔(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 .
In summation notation, this is just:

$\displaystyle \sum_{k=1}^n 2k-1 = 2\sum_{k=1}^n k -\sum_{k=1}^n 1 = 2\cdot\frac{n(n+1)}{2}-n = n^2+n-n = n^2$

If you want to do it your way, you can do what Gauss did...

$\displaystyle 1~~~~~~~+3~~~~~~~+...+2n-3+2n-1$
$\displaystyle \underline{2n-1+2n-3+...+3~~~~~~~+1~~~~~~~~~~+}$

So $\displaystyle 2S=2n\cdot n \implies S=n^2$

3. $\displaystyle 2k + 1 = (k+1)^2 - k^2$

Then solve it by method of differences

4. Thanks guys.