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Determine the sum of the following geometric series 1/32 + 1/16 + … + 256      

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alejandrogalarce | Student, Undergraduate | Valedictorian

Posted September 25, 2012 at 7:03 PM via web

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Determine the sum of the following geometric series 1/32 + 1/16 + … + 256

 

 

 

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted September 25, 2012 at 7:17 PM (Answer #1)

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The geometric series consists of the following terms, 1/32, 1/16, ... 256

The first term is a = 1/32.

From the first two terms the common ratio is r = `(1/16)/(1/32) = 2`

The nth term of a geometric series is `a*r^(n - 1)`

`256 = (1/32)*2^(n - 1)`

=> `2^(n - 1) = 256*32 = 2^13`

=> n = 14

The sum of the first 14 terms of the series is `(1/32)*(2^14 - 1)/(2 - 1) = 511.96875`

The sum of `1/32 + 1/16 + … + 256 = 511.96875`

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mun55 | Student, Grade 10 | eNoter

Posted September 25, 2012 at 7:54 PM (Answer #2)

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The geometric series consists of the following terms, 1/32, 1/16, ... 256

The first term is a = 1/32.

From the first two terms the common ratio is r = 

The nth term of a geometric series is 

=> 

=> n = 14

The sum of the first 14 terms of the series is 

The sum of 

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