How this summation is to 3?

  • Thread starter Buddy711
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  • #1
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Hi everyone.
I hardly remember the fomulas of summation of sequence.

I got this problem.

[tex]{\frac{1}{8}}\sum^{\infty}_{n=2}n({\frac{3}{4}})^{n-2}[/tex]

The result is 2.5.
How can I solve this problem?

Thanks all. :)
 
Last edited:

Answers and Replies

  • #2
286
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Assuming |r| < 1 then

[tex]\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}[/tex]

Differentiation both sides with respect to r gives:

[tex]\sum_{n=1}^{\infty} n \cdot r^{n-1} = \frac{1}{(1-r)^2}[/tex]

This should give you a push in the right direction.

(Warning: Be careful of your initial index.)

--Elucidus
 
  • #3
8
0
You suggested me very good approach.
However, the problem still remains,,,

my equation is n vs (n-2), not n vs (n-1)

Thanks!
 
  • #4
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,109
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Raising it to the power of n-2 instead of n-1 is just dividing it by 3/4. You should be able to find a way to modify your series so that you have an n-1 in the exponent
 
  • #5
8
0
Raising it to the power of n-2 instead of n-1 is just dividing it by 3/4. You should be able to find a way to modify your series so that you have an n-1 in the exponent
You are absolutely right.
I was so stupid.

Thank you ;-)
 

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