Trival question.

**Banned for attempted hacking** · May 9th 2009, 08:06 PM

$1 +3 +5 + ... (2n-1) = n^{2}$
How do we show this.

I realized that

This is the same thing as

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

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 ...

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

$k < n$

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

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

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

**redsoxfan325** · May 9th 2009, 09:31 PM

Originally Posted by ╔(σ_σ)╝

$1 +3 +5 + ... (2n-1) = n^{2}$
How do we show this.

I realized that

This is the same thing as

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

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 ...

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

$k < n$

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

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

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

In summation notation, this is just:

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

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

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

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

**simplependulum** · May 9th 2009, 09:43 PM

$2k + 1 = (k+1)^2 - k^2$

Then solve it by method of differences

**Banned for attempted hacking** · May 10th 2009, 05:40 AM

Thanks guys.

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