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Math Help - geometric sequences

  1. #1
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    Question geometric sequences

    show why
    (r^(n+1)-1)/(r-1) = the sum (between k=0 and n) of r^k = Sn (The n in Sn is lower than the S)

    he hints to multiply by r-1 or use long division

    then I'm supposed to use part to show why
    the sum K=0 to infinity = (1)/(1-r)

    i can graph the sequences and I know its true but I don't understand what to do tomake it work.

    thankyou,
    Cailtyn
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by horsejumper View Post
    show why
    (r^(n+1)-1)/(r-1) = the sum (between k=0 and n) of r^k = Sn (The n in Sn is lower than the S)

    he hints to multiply by r-1 or use long division
    (sum_{k=0 to n} r^k)(r-1) = sum_{k=0 to n} [r^{k+1} - r^k]

    .....................................= (r-1) + (r^2-r) + ... + (r^n-r^{n-1)) + (r^{n+1} - r^n)

    Now all but the -1 in the first term and the r^{n+1} in the last term cancel
    so:

    (sum_{k=0 to n} r^k)(r-1) = [r^{n+1}-1]

    or:

    sum_{k=0 to n} r^k = [r^{n+1}-1]/(r-1).

    then I'm supposed to use part to show why
    the sum K=0 to infinity = (1)/(1-r)
    From the first part we have:

    (sum_{k=0 to n} r^k) = [r^{n+1}-1]/(r-1)

    Now if |r|<1

    lim_{n->infty} (sum_{k=0 to n} r^k) = (sum_{k=0 to infty} r^k) = -1/(r-1) = 1/(1-r)

    as lim_{n->infty} r^{n+1} = 0.

    RonL
    Last edited by CaptainBlack; April 12th 2007 at 07:21 PM.
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  3. #3
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    Thankyou!!!!!!!!!!!!

    Thankyou soooo much!!!!!!!!!!!!!!!!!!!!!!!!!
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