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I am very confuse where to start and how to do it, I am very new to this kind of problems, would someone be able to guide me ?

Fred
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Volkan
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  • For starters, recognize that you are doing induction "on n." The inductive step is to prove that this holds true for n = k, then prove that it must follow for n = k +1. – The Chaz 2.0 Feb 16 '21 at 12:51
  • @TheChaz2.0 Yes thank you! I will try and see how far I can get ! – Volkan Feb 16 '21 at 13:06
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    Small typo: its $\frac{m^n -1}{m-1}$ not $\frac{m^{n -1}}{m-1}$ – Winther Feb 16 '21 at 13:11
  • And as stated, the result is true for all $m \neq 1$. – WA Don Feb 16 '21 at 13:17
  • A tip, so you can specify what you've tried and where you're stuck: with induction there are only two parts, the base case and the inductive step, so it's worth saying which troubled you. – J.G. Feb 16 '21 at 15:01

2 Answers2

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The inductive step would be: $\color{blue}{1+m+\dots+m^{n-1}}+m^n=\color{blue}{\dfrac{m^n-1}{m-1}}+m^n=\dfrac{m^n-1}{m-1}+\dfrac{m^{n+1}-m^n}{m-1}=\dfrac{m^{n+1}-1}{m-1}$.

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Base step: $n=1$ then $\frac{m^1-1}{m-1}=1$

Induction hypothesis: suppose it is valid for $n = k$.

Inductive step: Let \begin{align} (1+m+m^2+\ldots+m^{k-1})+m^{(k-1)+1}&=\frac{m^k-1}{m-1}+m^k\\ &=\frac{m^k-1+m^{k+1}-m^k}{m-1}\\ &=\frac{m^{k+1}-1}{m-1}\\ \end{align} with this you can already argue

logarithm
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