Finding the formula for summation.

  • Thread starter NATURE.M
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  • #1
NATURE.M
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Homework Statement



Find a formula for

[itex]\sum (2i-1)[/itex] =1+3+5+...+(2n-1)

Homework Equations





The Attempt at a Solution



[itex]\sum(2i-1)[/itex]=(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 ?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
statdad
Homework Helper
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Maybe start with
[tex]
\sum_{i=1}^n \left(2i-1\right) = 2\sum_{i=1}^n i - \sum_{i=1}^n 1
[/tex]

and simplify?
 
  • #3
arildno
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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
NATURE.M
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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
arildno
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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!
:smile:
 

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