- #1
esvion
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I have to find the sum of [tex]\sum9(2/3)^n[/tex] 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?
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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