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Math Help - Sequence

  1. #1
    Junior Member
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    Sequence

    How would I do this problem?

    Use the formula a[sub-1]+a[sub-1]r+a[sub-1]r[squared]+...+a[sub-1]r[n-1 power]=

    a[sub-1]* 1-r[nth power]/1-r

    for the nth partial sum of a geometric sequence to find the sum 1+2/3+4/9+8/27+...+64/729.

    possible answers:
    701/2187
    2059/729
    2187/701
    2187/729
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by gretchen View Post
    How would I do this problem?

    Use the formula a[sub-1]+a[sub-1]r+a[sub-1]r[squared]+...+a[sub-1]r[n-1 power]=

    a[sub-1]* 1-r[nth power]/1-r

    for the nth partial sum of a geometric sequence to find the sum 1+2/3+4/9+8/27+...+64/729.

    possible answers:
    701/2187
    2059/729
    2187/701
    2187/729
    The summation formula for a finite geometric series is:

    a+a r+a r^2+...+a r^(n-1) = a [1-r^n]/[1-r]

    in the example given a=1, r=(2/3), and n=7 (as 64/729=(2/3)^6).

    So the sum of the given series is:

    1 [1-(2/3)^7]/[1-2/3]=2059/729

    RonL
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