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I am having difficulties in solving the expression $1+x+x^2+\cdots+x^{n-1} =\frac{1-x^n}{1-x}$, $n\in \mathbb{N}$, using the induction method. How is the inductive step taken in this case?

Angina Seng
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Felipe Duda
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  • This should be as straightforward as an induction proof can be. What is the obstacle? Add $x^n$ to both sides. What happens to the right-hand side? – Mark Viola Nov 03 '19 at 18:06
  • The obstacle is how to take the inductive step. The induction base is taken properly: For $n=1$, $1=\frac{1-x^1}{1-x}=1$

    The problem is how to take the next step, how to prove for $n+1$.

     – Felipe Duda Nov 03 '19 at 18:08
  • This question has been answered here:https://math.stackexchange.com/questions/1502997/induction-proof-dealing-with-geometric-series – NoChance Nov 03 '19 at 18:08
  • @felipeduda Read my entire comment. Answer the question embedded therein. You have the answer that Ennar posted, which you stated you understood. – Mark Viola Nov 03 '19 at 18:21
  • I did! Thank you! – Felipe Duda Nov 03 '19 at 18:39

2 Answers2

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Assume that the formula holds for some $n\in\mathbb N$. Then

$$\underbrace{1+x+\ldots + x^{n-1}}_{\text{formula holds for this part}}+x^n = \frac{1-x^n}{1-x} + x^n.$$

Can you finish now?

Ennar
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Assume that $1+x + \dots + x^{n-1} = \frac{1-x^n}{1-x}$.

Then

$$1+x+ \dots + x^{n} = (1+ x + \dots + x^{n-1}) + x^n = \frac{1-x^n}{1-x} + x^n$$ $$= \frac{1-x^n}{1-x} + \frac{x^n(1-x)}{1-x} = \frac{1-x^n + x^n - x^{n+1}}{1-x}= \frac{1-x^{n+1}}{1-x}$$

J. De Ro
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