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Math Help - Writing a recursive definition for Sn

  1. #1
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    Writing a recursive definition for Sn

    Find a formula for tn, the nth term of the series. Then give a recursive definition for Sn, the sum of n terms of the series.

    1) 8+12+18+27+....

    2) 50+47+44+41+....

    I know how to find the tn
    tn = tn-1 * (3/2)
    tn = 53-3n

    My teacher gave us the answers
    1. Sn = Sn-1 + 8(3/2)^(n-1)
    2. Sn = Sn-1 +53-3n

    Can somebody explain to me why these two are the answers for the recursive definition of Sn and how to write one? Thanks!
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Writing a recursive definition for Sn

    Quote Originally Posted by njuice8 View Post
    Find a formula for tn, the nth term of the series. Then give a recursive definition for Sn, the sum of n terms of the series.

    1) 8+12+18+27+....

    2) 50+47+44+41+....

    I know how to find the tn
    tn = tn-1 * (3/2)
    tn = 53-3n

    My teacher gave us the answers
    1. Sn = Sn-1 + 8(3/2)^(n-1)
    2. Sn = Sn-1 +53-3n

    Can somebody explain to me why these two are the answers for the recursive definition of Sn and how to write one? Thanks!
    Hi njuice8!

    A series is the sum of the terms.
    So Sn is the sum of the first n terms.
    Or put differently, the sum of the first (n-1) terms plus the nth term.
    In formula form:

    S_n = t_1 + t_2 + ... + t_{n-1} + t_n

    S_{n-1} = t_1 + t_2 + ... + t_{n-1}

    S_n = S_{n-1} + t_n

    If you fill in the expression for t_n you'll get both of the recursive formulas.
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  3. #3
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    Re: Writing a recursive definition for Sn

    The first one is a GP with the first term a = 8 and common ration r = 3/2 { you are right}
    n^th term will be given by ar^n-1 and sum of n terms Sn = ( r^n-1)/(r-1)
    The second one is an AP with the first term a = 50 and common difference d = -3
    The n^th term is given by Tn = a + ( n-1) d
    and sum of the n terms Sn = n/2 [ 2a + ( n-1 ) d]
    Plug in the values and get the answers.
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