# How this summation is to 3?

1. Oct 8, 2009

### Buddy711

Hi everyone.
I hardly remember the fomulas of summation of sequence.

I got this problem.

$${\frac{1}{8}}\sum^{\infty}_{n=2}n({\frac{3}{4}})^{n-2}$$

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

Thanks all. :)

Last edited: Oct 9, 2009
2. Oct 8, 2009

### Elucidus

Assuming |r| < 1 then

$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$

Differentiation both sides with respect to r gives:

$$\sum_{n=1}^{\infty} n \cdot r^{n-1} = \frac{1}{(1-r)^2}$$

This should give you a push in the right direction.

(Warning: Be careful of your initial index.)

--Elucidus

3. Oct 9, 2009

### Buddy711

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. Oct 9, 2009

### Office_Shredder

Staff Emeritus
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. Oct 9, 2009

### Buddy711

You are absolutely right.
I was so stupid.

Thank you ;-)