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I'm reading through a number theory text and the following equivalence is used in a proof. It looks kind of like a binomial expansion, but not. I don't understand why this is true.

$a^n - 1= (a-1)(a^{n-1}+a^{n-2}+ \cdots + 1)$

Edit:

For the record, I can multiply. I'm curious to know why generally this is true, and how you get from $a^n -1$ to the above.

Okay, really what I'm asking, is how do you uncollapse a geometric series? Since that's what the right side of the equation is right? Is there a general method to do this?

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If $a=10$ then you could think about $999\ldots999 = 9 \times 111\ldots111$. Then note that there is nothing special about $10$ here. –  Henry Feb 23 '12 at 23:32
    
Okay, this makes this make sense. Really, It's just like we can say we're subtracting 1 in some arbitrary base a from a power of that base. Can you re-write your comment as an answer? I will accept it. –  Josh Infiesto Feb 23 '12 at 23:36

4 Answers 4

up vote 1 down vote accepted

If $a=10$ then you could think about $999\ldots 999=9\times 111\ldots 111$. Then note that there is nothing special about $10$ here.

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It's just cancellation, $$ a \cdot (a^{n-1}+a^{n-2}+ \cdots + 1) = a^n+a^{n-1}+ \cdots + a, $$ subtract off $$ 1 \cdot (a^{n-1}+a^{n-2}+ \cdots + 1) $$ and you get $$ a^n - 1. $$

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$\rm Hint:\:\ (1,1,1,\ldots,1,1,0)\ \:[=\:$ leftshift of vector below]
$\rm minus\ \ (0,1,1,\ldots,1,1,1)$
$\rm equals\:\ (1,0,0,\ldots,0,0,-\!1)$

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Try multiplying the terms to get rid of the brackets. You'll find that many terms vanish....

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