Thanks!!
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$\displaystyle \frac{{4^n }} {{3^{n - 1} }} = \frac{4} {4} \cdot \frac{{4^n }} {{3^{n - 1} }} = 4 \cdot \frac{{4^{n - 1} }} {{3^{n - 1} }}$ The conclusion follows.
Originally Posted by Krizalid $\displaystyle \frac{{4^n }} {{3^{n - 1} }} = \frac{4} {4} \cdot \frac{{4^n }} {{3^{n - 1} }} = 4 \cdot \frac{{4^{n - 1} }} {{3^{n - 1} }}$ The conclusion follows. Thanks but I still have a couple of questions 1) How did you know to pull out a 4 in step 2? 2) How are you able to get n-1 in $\displaystyle \\{{4^{n - 1} }}$ ?
It's all based on the following properties: $\displaystyle a^m:a^n=a^{m-n}$ & $\displaystyle \frac{{a^k }} {{b^k }} = \left( {\frac{a} {b}} \right)^k$ By the way, this has nothing to do with "Calculus".
Originally Posted by Krizalid By the way, this has nothing to do with "Calculus". Series
Your question was special. In that case you should've posted the full question.
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