Finding the formula for summation.

[NATURE.M]

Homework Statement

Find a formula for

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

Homework Equations

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 ?

 

[statdad]

Maybe start with
<br /> \sum_{i=1}^n \left(2i-1\right) = 2\sum_{i=1}^n i - \sum_{i=1}^n 1<br />

and simplify?

 

[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?

 

[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.

 

[arildno]

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.

Yup!