1. ## 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?

2. Try induction!

3. 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.

4. Well I can give you the identity $\sum_{j=1}^nj=\frac{n(n+1)}{2}$. See if you can solve the problem using this, and then if you wish we will look at how to prove it.

5. 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.

6. Good. Be careful with your notation - be as clear as possible, don't put more than one equal sign in your expression!

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

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

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

and add it to the same sum, written backwards :

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

Add them vertically, term by term. What do you get?

7. 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

8. Yes; I have showed you how to do that, but I've used an identity which we have not proved, namely $1+2+...+n=n(n+1)/2$. I was suggesting a proof of the latter identity.

9. 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

10. Yes, that is what you have to do. Read my post above :

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

11. 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!