# Show $1 + a + a^2 + a^3 + \ldots + a^n = \frac{a^{n+1}-1}{a-1}$ by induction

How can we show by mathematical induction that the following holds for $n \ge 0$ and $a \ne 1$?

$$1 + a + a^2 + a^3 + \ldots + a^n = \frac{a^{n+1}-1}{a-1}$$

I understand the principle of mathematical induction, but I've no idea how to apply it here. I know in general I have to prove it for $n = 1$ and then assume $n = k$ is correct. Then prove $n = k+1$ is true. But what about $a$?

I've watched a load of YouTube videos on the subject, and I've read a few questions here but it's not helping. The videos make sense while I'm watching them, but I don't know how to apply it. This question appeared on my discrete math exam last week. I did not do well. I think I'm missing something fundamental in my understanding of this subject.

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If $n=0$, we must prove $1=\frac{a-1}{a-1}$ which is trivial.

Assume that it holds for $n=k$, that is: $$1+a+...+a^k=\frac{a^{k+1}-1}{a-1}$$ We must prove that it holds for $n=k+1$ or in other words that $$1+a+...+a^k+a^{k+1}=\frac{a^{k+2}-1}{a-1}$$ But this is simple to prove with our assumption: $$1+a+...+a^k+a^{k+1}=\frac{a^{k+1}-1}{a-1}+a^{k+1}=\frac{a^{k+1}-1+a^{k+2}-a^{k+1}}{a-1}=\frac{a^{k+2}-1}{a-1}$$

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Note that he is supposed to start at $n=0$ (which is even easier to start with), not at $n=1$. –  Hagen von Eitzen Dec 17 '12 at 16:47
Thank you, I will study this and come back to you. –  bot_bot Dec 17 '12 at 16:48
Hi Nameless, thank you for your help. –  bot_bot Jan 9 at 9:27
Let $P(n)$ be the statement that $1+a+...+a^n=\frac{a^{n+1}-1}{a-1}$. Since $1=\frac{a-1}{a-1}$, $P(0)$ is true.
Suppose $P(k)$ is true for some non-negative integer $k$. Then $1+a+...+a^k=\frac{a^{k+1}-1}{a-1}$, so that $1+a+...+a^{k+1}=\frac{a^{k+1}-1}{a-1}+a^{k+1}=\frac{a^{k+1}-1}{a-1}+\frac{a^{k+2}-a^{k+1}}{a-1}=\frac{a^{k+2}-1}{a-1}$, so that $P(k+1)$ is true.
Hence we have $P(n)$ for all non-negative integers $n$.
Note that in this example, you are asked to start at $n=0$ and not $n=1$ and thus prove the statement for all non-negative integers and not just positive integers. Also note that here $a\neq 1$ is fixed and we are doing induction on $n$ and not on $a$.
Hi Jasper, did you lose a $-1$ term in the final numerator? –  Amzoti Dec 17 '12 at 16:54