Sum of geometric series starting at n=1

[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?

 

[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?

[tex]\sum_{n=0}^\infty a r^n= \frac{a}{1- r}[/itex]<br /> <br /> but <br /> $$\sum_{n=1}^\infty a r^n= r \sum_{n=0}^\infty a r^n$$<br /> where I have factored out an "r" to reduce the first index to 0,<br /> and<br /> $$\sum_{n=1}^\infty a r^n= \left(\sum_{n=0}^\infty a r^n\right) - a$$<br /> where I have subracted off the $ar^0$ term.<br /> <br /> In particular, with a= 9, r= 2/3, <br /> $$\sum_{n=1}^\infty 9 (2/3)^n= (2/3)\sum_{n=0}^\infty 9 (2/3)^n$$<br /> $$= \frac{2}{3}\frac{9}{1- 2/3}= \frac{6}{1- 2/3}= 18$$<br /> <br /> or <br /> $$\sum_{n=1}^\infty 9(2/3)^n= \left(\sum_{n=0}^\infty 9 (2/3)^n\right)- 9$$<br /> $$= \frac{9}{1- 2/3}- 9= 27- 9= 18$$[/tex]

 

[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