# Finding the formula for summation.

1. Sep 14, 2013

### NATURE.M

1. The problem statement, all variables and given/known data

Find a formula for

$\sum (2i-1)$ =1+3+5+...+(2n-1)

2. Relevant equations

3. The attempt at a solution

$\sum(2i-1)$=(1+2+3+...+2n)-(2+4+6+...+2n)
=(1+2+3+...+2n)-n(n+1)

I'm unsure what to do with 1+2+3+..+2n ?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 14, 2013

$$\sum_{i=1}^n \left(2i-1\right) = 2\sum_{i=1}^n i - \sum_{i=1}^n 1$$

and simplify?

3. Sep 14, 2013

### arildno

statdad's suggestion is one way of doing this. Alternatively, you might spot a pattern by looking at the partial sums:
1=?
1+3=?
1+3+5=?

Once you spot the pattern, how could you go about proving whether it is actually universally true, or a mere fluke?

4. Sep 14, 2013

### NATURE.M

I think I fugured it out lol.

The pattern is n^2.
So then,

(1+2+3+...+2n)-n(n+1) = (2n(2n+1))/2 -n(n+1)=n(2n+1)- n(n+1) =2n^2+n-n^2-n=n^2
So the general formula is n^2.

5. Sep 14, 2013

Yup!