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Thread: Proof By Induction

  1. #1
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    Proof By Induction

    Hey, I was wondering if someone could help me with a question I have.

    Prove by induction

    x^n-1 + x^n-2 + .... + x + 1 = (x^n)-1 / x - 1

    Thank you
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  2. #2
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    Re: Proof By Induction

    You need grouping symbols.

    Prove by induction for  n \ge 1.

    x^(n - 1) + x^(n - 2) + ... + x + 1 = (x^n - 1)/(x - 1)

    Have you done the base case, n = 1?
    Last edited by greg1313; Nov 8th 2017 at 08:31 PM.
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  3. #3
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    Re: Proof By Induction

    Quote Originally Posted by jkd98 View Post
    Prove by induction
    x^n-1 + x^n-2 + .... + x + 1 = (x^n)-1 / x - 1
    Thank you
    It is very clear that this is the case if $n=1$

    So suppose that $J>1~\&~\dfrac{{{x^J} - 1}}{{x - 1}} = \sum\limits_{k = 0}^{J - 1} {{x^k}}$

    $ \begin{align*}\frac{{{x^{J + 1}} - 1}}{{x - 1}} &= {x^{J}} + \frac{{{x^J} - 1}}{{x - 1}}\\&= {x^{J + 1}} + \sum\limits_{k = 0}^J {{x^k}}\\&= \sum\limits_{k = 0}^{J + 1} {{x^k}} \end{align*}$
    Last edited by Plato; Nov 8th 2017 at 08:41 PM.
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  4. #4
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    Re: Proof By Induction

    \begin{align*}x^n + x^{n-1} + x^{n-2} + \ldots + 1 &= x^n + (x^{n-1} + x^{n-2} + \ldots + 1) \\ &=x^n + \frac{x^n - 1}{x-1} &\text{by the induction hypothesis} \\ &= \frac{x-1}{x-1}x^n + \frac{x^n - 1}{x-1}\end{align*}
    I'm sure you can finish from there.
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