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this step in the proof is confusing me:

$\sum_1^\infty {\frac{4^{n}}{3^{n-1}}}\qquad \longrightarrow \qquad\sum_1^\infty 4\left(\frac43\right)^{n-1}$

please explain how/why this happened?

cheers,

gregg

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    Moron, Jonas thanks for clearing that up - newbie to the site and calculus so I appreciate your help.2010-12-03

2 Answers 2

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The $n^\text{th}$ term was rewritten by pulling out a factor of $4$ from the numerator. Maybe seeing a couple of extra steps will help:

$\frac{4^n}{3^{n-1}}=\frac{4\cdot4^{n-1}}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}=4\left(\frac{4}{3}\right)^{n-1}$

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    (Users with less than 15 points can't vote, and one answer with positive vote total suffices to remove a question from the Unanswered list.)2010-12-03