1. Oct 3, 2011

### rlpp

1. The problem statement, all variables and given/known data
Find the formula
Σ[n, i =1](2i - 1) = 1+3+5+...+(2n-1)

2. Relevant equations

3. The attempt at a solution
When I look at the back of the book , I got this solution which is

1, Sigma(2i - 1) = 1+3+5+...+(2n-1)

2, = 1+2+3+...+2n-2(1+..+n)

3, = ((2n)(2n+1)) /2 - n(n+1)

4, n^2

What i don't get is what happened from step 1 to 2 and 2 to 3?
It seems that there are hidden steps there that were not written down and therefore and I am really confused about this solution.
Thank you guys very much.

2. Oct 3, 2011

### Dick

Your given sum is the sum of all of the odd numbers up to 2n. Step 2 is pretty easy. He just rewrote that as the sum of ALL of the numbers up to 2n minus the sum of all of the even numbers up to 2n. The trick you might be missing is that the sum of all of the numbers up to n, Sigma(i) for i=1 to n is given by a formula. It's n*(n+1)/2. There's at least a couple of ways to prove that. Do you know either one?

3. Oct 4, 2011

### rlpp

Thank you very much, I am still a little confused but I will work with this formula. Spivak's calculus is so confusing lol.