Solving Summations with Modified Exponents

[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. :)

 

[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

 

[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!

 

[Office_Shredder]

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

 

[Buddy711]

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 ;-)