# Sum of geometric series

#### esvion

I have to find the sum of $$\sum9(2/3)^n$$ and I get a/1-r where a=9 and r=2/3.....but I know a=6 and not 9. Can someone point out to me what I am doing wrong? The sum is from n=1 to infinity.

Thanks!

EDIT: I am thinking I take a(1) which is 6 as the a in a/(1-r), is this correct?

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#### foxjwill

I am thinking I take a(1) which is 6 as the a in a/(1-r), is this correct?
Yep!

#### HallsofIvy

I have to find the sum of $$\sum9(2/3)^n$$ and I get a/1-r where a=9 and r=2/3.....but I know a=6 and not 9. Can someone point out to me what I am doing wrong? The sum is from n=1 to infinity.

Thanks!

EDIT: I am thinking I take a(1) which is 6 as the a in a/(1-r), is this correct?
You left out important information! What is the beginning index for your sum?

$$\sum_{n=0}^\infty a r^n= \frac{a}{1- r}[/itex] but [tex]\sum_{n=1}^\infty a r^n= r \sum_{n=0}^\infty a r^n$$
where I have factored out an "r" to reduce the first index to 0,
and
$$\sum_{n=1}^\infty a r^n= \left(\sum_{n=0}^\infty a r^n\right) - a$$
where I have subracted off the $ar^0$ term.

In particular, with a= 9, r= 2/3,
$$\sum_{n=1}^\infty 9 (2/3)^n= (2/3)\sum_{n=0}^\infty 9 (2/3)^n$$
$$= \frac{2}{3}\frac{9}{1- 2/3}= \frac{6}{1- 2/3}= 18$$

or
$$\sum_{n=1}^\infty 9(2/3)^n= \left(\sum_{n=0}^\infty 9 (2/3)^n\right)- 9$$
$$= \frac{9}{1- 2/3}- 9= 27- 9= 18$$

#### annoymage

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

hmm, may i know how to factorise that r please?

#### HallsofIvy

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

hmm, may i know how to factorise that r please?
It's just the "distributive law": $ar+ ar^2+ ar^3+ \cdot\cdot\cdot= r(a+ ar+ ar^2+ \cdot\cdot\cdot$

#### annoymage

owh, thank you very much

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