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Thread: 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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    $\displaystyle \frac{a+b(1+n+n^2+\cdots+n^{k-1})}{n^k}$
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  3. #3
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    Hello, Pinky9!

    $\displaystyle a$ add $\displaystyle b$, divide by $\displaystyle n$,
    . . . add $\displaystyle b$, divide by $\displaystyle n$,
    . . . add $\displaystyle b$, divide by $\displaystyle n$ . . . recurring.

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

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

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

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

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


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

    . . $\displaystyle \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: .$\displaystyle \frac{1-n^k}{1-n}$ . for $\displaystyle n \neq 1$
    $\displaystyle \text{Hence: }\:u_k \;=\;\frac{1+b\,\frac{1-n^k}{1-n}}{n^k} $

    $\displaystyle \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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