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Find the sum of the first six terms in the geometric sequence 1/2, 1/8, 1/16, .... Express your answer as a common fraction.

 Jun 12, 2021

Best Answer 

 #1
avatar+517 
+2

I think you missed a term, the GP should be : 1/2, 1/4, 1/8, 1/16,...

⇒ a = 1/2

    r = 1/2

    n = 6

 

\({S}_{n}={a(1-r^n)\over 1-r}\)

      \(= {{1\over 2}(1- {1\over 2^6})\over {1- {1\over 2}}}\)

      \(=1- {1\over 2^6}\)

      \(={63 \over 64}\)

 

\({S}_{6}= {63\over 64}\)

 Jun 12, 2021
 #1
avatar+517 
+2
Best Answer

I think you missed a term, the GP should be : 1/2, 1/4, 1/8, 1/16,...

⇒ a = 1/2

    r = 1/2

    n = 6

 

\({S}_{n}={a(1-r^n)\over 1-r}\)

      \(= {{1\over 2}(1- {1\over 2^6})\over {1- {1\over 2}}}\)

      \(=1- {1\over 2^6}\)

      \(={63 \over 64}\)

 

\({S}_{6}= {63\over 64}\)

amygdaleon305 Jun 12, 2021

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