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# Help Need

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Find the sum of the first five terms in the geometric sequence $$\frac13,\frac19,\frac1{27},\dots$$. Express your answer as a common fraction.

Jan 5, 2021

#2
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CPhill I got something different

This is a finite geometric series with first term $$1/3$$ and common ratio $$1/3$$. There are five terms, so the sum of this series is $$\frac{\frac13\left(1-\left(\frac13\right)^5\right)}{1-\frac13} = \boxed{\frac{121}{243}}$$.

Jan 5, 2021

#1
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Sum of a geometric series

S =  first term  (  1 -  common ratio ^n)   / ( 1  - common ratio)     where n =  the  number of terms we are  summing

The common ratio  is    (1/9) / (1/3)  =  3/9  = 1/3

So

S =   ( 1/3)  ( 1 - (1/3)^3 ) / ( 1 - 1/3)   =

(1/3)  ( 1 - 1/27) / ( 2/3)   =

(1/3) (3/2)  * ( 26/27 )   =

(1/2) ( 26/27)  =

26/54  =

13/27

Jan 5, 2021
#2
+303
+1

CPhill I got something different

This is a finite geometric series with first term $$1/3$$ and common ratio $$1/3$$. There are five terms, so the sum of this series is $$\frac{\frac13\left(1-\left(\frac13\right)^5\right)}{1-\frac13} = \boxed{\frac{121}{243}}$$.

hihihi Jan 5, 2021
#3
+117725
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OOOPS!!!...misread the problem....I just did the first three terms....your answer is  correct  !!!

THX for the  correction  !!

CPhill  Jan 5, 2021
#4
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This ok

Jan 5, 2021