# How this summation is to 3?

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:

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

You suggested me very good approach.
However, the problem still remains,,,

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

Thanks!

Office_Shredder
Staff Emeritus