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What is the formula of:

$$a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$$

Any ideas?

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Hint: For $a=1$ this is simple. Otherwise, what happens when you multiply it by $a-1$?

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    $\begingroup$ +1. What happens? We watch the dominoes falling... $\endgroup$ – Did Apr 10 '13 at 15:58
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See

http://en.wikipedia.org/wiki/Geometric_series

it is called a gemoetric series and it is a standard result.

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If $a=1,a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}=1+1+\cdots$ up to $(n+1)$ terms hence $=n+1$

Else let $S=a^{0} + a^{1} + a^{2} + ... + a^{n-1} + a^{n}$

So, $a\cdot S=a^{1} + a^{2} + a^{3} + ... + a^{n} + a^{n+1}$

So, $S(a-1)=a^{n+1}-1$

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$$a^{n}-1=(a-1)(a^{n-1}+\cdots+1)\implies a^{n-1}+\cdots+1=\dfrac{a^{n}-1}{a-1}$$ where $a \neq 1$

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    $\begingroup$ Check your algebra. $\endgroup$ – leo Apr 10 '13 at 16:15
  • $\begingroup$ @leo: Thanks. I didn't notice that. $\endgroup$ – Inceptio Apr 10 '13 at 16:31

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