How do I show that the sum from {i=1} to {n} (2i-1) is equal to n^2 for all natural numbers n?
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 $\displaystyle \sum_{j=1}^n(2j-1)=2\left\{n(n+1)/2\right\}-n=n^2+n-n=n^2$.
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?
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