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Math Help - number properties

  1. #1
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    number properties

    a add b divide by n, add b divide by n, add b divide by n.....recurring?

    Teachers doesn't ask a specific question but requires an investigation into this...anyone have an idea where to start? I have tried using different types of denominators as well as lower or higher numerators but no recurring pattern emerges.
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  2. #2
    tah
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    \frac{a+b(1+n+n^2+\cdots+n^{k-1})}{n^k}
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  3. #3
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    Hello, Pinky9!

    a add b, divide by n,
    . . . add b, divide by n,
    . . . add b, divide by n . . . recurring.

    With no instructions, I will assume we want the {\color{blue}k^{th}} term of the sequence.
    Crank out the first few terms and look for a pattern . . .
    . . Note that n \neq 0

    . . u_1 \:=\:\frac{a+b}{n}

    . . u_2 \:=\:\frac{\frac{a+b}{n} + b}{n} \:=\:\frac{a+b+bn}{n^2}

    . . u_3 \:=\:\frac{\frac{a+b+bn}{n} + b}{n} \;=\;\frac{a+b+bn + bn^2}{n^3}

    . . u_4 \:=\:\frac{\frac{a+b+bn+bn^2}{n} + b}{n} \:=\:\frac{a+b+bn+bn^2+bn^3}{n^4}


    The k^{th} term seems to be: . u_k \;=\;\frac{a+b + bn + bn^2 + \hdots + bn^{k-1}}{n^k}

    . . \text{and we have: }\:u_k \;=\;\frac{a + b\overbrace{(1 + n + n^2 + \hdots + n^{k-1})}^{\text{geometric series}}}{n^k}


    The geometric series has the sum: . \frac{1-n^k}{1-n} . for n \neq 1
    \text{Hence: }\:u_k \;=\;\frac{1+b\,\frac{1-n^k}{1-n}}{n^k}

    \text{Therefore: }\:\boxed{u_k \;=\;\frac{(1-n)a + (1-n^k)b}{n^k(1-n)}}

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  4. #4
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    Thank you so much - I should have thought of that!
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