The sume of the infinite geometric series 3/2+9/16+27/128+81/1024 + ….. is?

How do I do this problem ????

2 Answers

  • 3/2+9/16+27/128+81/1024 + .....

    Let S be the sum of the given series. Then

    S = (3/2) + (9/16) + (27/128) + (81/1024) + .......... inf.

    Here the common ratio (r) = Any term divided by the term that follows it.

    For example choose the term (81/1024) the following term is (27/128) hence -

    Common ratio (r) = (81/1024) divided by (27/128)

    => r = 3/8 This common ratio is less than one, hence the sum to infinite terms can be found out. ( Note that exact sum of a GP having infinite number of terms can be obtained iff, ' r ' is a fraction.)

    Hence the required sum : -

    S = [ a / ( 1 - r ) ] , where a is first term of the series,

    => S = [ 3/2 divided by ( 1 - 3/8 ) ]

    => S = ( 3/2 ) divided by ( 5/8 )

    => S = 12/5 ..................................... Answer

    PKT

    Source(s): Its me only at <[email protected]>
  • a = 3/2

    r = 3/8

    S = a(1 - r^n)/(1 - r)

    = (3/2)[1 - (3/8)^infinite]/[1 - 3/8]

    (3/8)^infinite = 0 since the value becomes smaller and smaller as the power is increased.

    So

    S = a / (1 - r) = 3/2 * 8/5 = 24/10

    Hope this helps.

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