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Math Help - Nth term in a geometric series

  1. #1
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    Nth term in a geometric series

    Nth term in a geometric series-math-help.jpgSo for this I have calculated what I believe to be the correct few terms of Q and P Q0=75, Q1=84, Q2=94.08. Always increasing by 12% of what the previous function was. I can't understand why my formula didn't work. I also tried using Qn=Q0*Rn-1 but that provided an answer for n=0that was smaller than 75.

    *Edit, I understand that I cant use n-1 because we're starting from n=0 and not n=1 so shouldn't it just be Qn=Q0*Rn?
    Last edited by bikerboy2442; March 21st 2013 at 11:40 PM.
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: Nth term in a geometric series

    If I understand the question correctly, then
    Q_0 = 75
    Q_n= (.12)Q_{n-1} + 75, in other words, what remains in the system on the previous day plus the new dose of 75mg per day.

    EDIT:// this appears right in terms of a recursive equation, now all you would need to do is simplify to closed form.
    Last edited by MacstersUndead; March 22nd 2013 at 12:19 AM.
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