# Proving Identities

• Oct 28th 2009, 12:13 PM
Velvet Love
Proving Identities
How do I show that the sum from {i=1} to {n} (2i-1) is equal to n^2 for all natural numbers n?
• Oct 28th 2009, 12:26 PM
Bruno J.
Try induction!
• Oct 28th 2009, 12:28 PM
Velvet Love
I don't know what that is. XD
We haven't learned it yet.
When I look at the formulas and rules, I find nothing that helps at all.
• Oct 28th 2009, 12:36 PM
Bruno J.
Well I can give you the identity $\displaystyle \sum_{j=1}^nj=\frac{n(n+1)}{2}$. See if you can solve the problem using this, and then if you wish we will look at how to prove it.
• Oct 28th 2009, 12:41 PM
Velvet Love
Alright. Now i've gotten the original sum down to this:

2* sum from {i=1} to {n} i = (n(n+1))/2 - sum from {i=1} to {n} 1 = n.

Now I think I have to prove it, but I don't know how.
• Oct 28th 2009, 12:47 PM
Bruno J.

So you have $\displaystyle \sum_{j=1}^n(2j-1)=2\left\{n(n+1)/2\right\}-n=n^2+n-n=n^2$.

Now, to prove that $\displaystyle 1+2+...+n=n(n+1)/2$, consider

$\displaystyle S=1+2+...+n$

and add it to the same sum, written backwards :

$\displaystyle S=n+(n-1)+..+1$.

Add them vertically, term by term. What do you get?
• Oct 28th 2009, 12:57 PM
Velvet Love
I'm confused as to what you are asking.
I have to prove that all of that equals n^2, for all natural numbers n
• Oct 28th 2009, 12:59 PM
Bruno J.
Yes; I have showed you how to do that, but I've used an identity which we have not proved, namely $\displaystyle 1+2+...+n=n(n+1)/2$. I was suggesting a proof of the latter identity.
• Oct 28th 2009, 01:04 PM
Velvet Love
Oh ok. I have one question. Since the problem has 2 in front of the sigma notation in the first term, could i cancel the two out to get sum from {i=1} to {n} n^2 + n - sum from {i=1} to {n} n?
And then I could cancel one n out, which would give me
sum from {i=1} to {n} n^2?
I think that is how you prove it, but I cannot be sure
• Oct 28th 2009, 01:26 PM
Bruno J.
Yes, that is what you have to do. Read my post above :

Quote:

Originally Posted by Bruno J.
So you have $\displaystyle \sum_{j=1}^n(2j-1)=2\left\{n(n+1)/2\right\}-n=n^2+n-n=n^2$.

• Oct 28th 2009, 01:33 PM
Velvet Love
Oh ok. I looked at that equation and I just didn't think. Thank you though! You pointed me in the right direction, and gave me the answer, but I didn't realize you gave me the answer, and I got it myself.
Thanks!