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Hello StackExchange world,

While doing a proof, I encountered: $ = S(n-1)+3^n-3^{n-1} $

I am focused on the latter part, with the powers of n, apparently it reduces after factoring to:

$ 2*3^{n-1} $

and I have no idea why. I tried thinking of n as a concrete number, and then n-1 the one before that, and I still can't wrap my mind around it. I am hoping someone can enlighten me here. Thank you all for the time and effort.

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    $3^n=3*3^{n-1}$ and you've subtracted another $3^{n-1}$ so...2012-07-11

2 Answers 2

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$3^n-3^{n-1}=3.3^{n-1}-3^{n-1}=3^{n-1}(3-1)=2.3^{n-1}.$ I think you thought it as $3^{n}-3(n-1)$,which is not right.

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    Wow thank you so much. I need to learn these little formatting tricks, would of made the algebra so much simpler.2012-07-11
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It's actually pretty simple. It suffices to factor out $3^{n-1}$ and note that $3^{n-1} \times 3 = 3^{n}$.

$ 3^{n} - 3^{n-1} = 3^{n-1}(3 - 1) = 2 \times 3^{n-1} $

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    This [pdf](http://www.freeteachingresources.ca/Exponent%20laws%20and%20factoring%20tips.pdf) seems to cover most of the basic laws for exponentiation. There's no magic here, whenever you see similar numbers/letters, try factoring them out. But the best piece of advice anyone could give is for you to get a good night's sleep. It does miracles.2012-07-11