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I need to convert the following series into a form that works for the equation $$\frac{a}{1-r}$$ so that I can calculate its sum. But the relevant laws of exponents are eluding me right now.

$$\sum_{n=1}^{\infty}\left(\frac{4}{10}\right)^{3n-1}$$

How do I get the 3 out of the exponent?

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Out of curiosity, why is this tag "precalculus" in use? As far as I know it's a strange word only used in the USA, and in any case is an umbrella term for some very disparate subfields of basic mathematics. Most Europeans (and I believe I speak for Asians too) would have no clue what it means. –  Noldorin May 27 '13 at 20:34
    
Maybe it is North American, I don't know. To me it always meant things immediately preceding the introduction to calculus... So stuff to do with trig, quadratic equations, etc... Google says "A course in mathematics that prepares a student for calculus." –  agent154 May 27 '13 at 20:36
    
Fair enough. But here in Britain for example, we don't have a specific course that prepares us for calculus. (Indeed, we generally start studying calculus at a younger age, and the age is even younger in parts of Europe.) I suppose we don't even have a rigorously delineation of fields of mathematics when we study, so it wouldn't make much sense to us anyway. –  Noldorin May 27 '13 at 21:34

2 Answers 2

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Hint: $$\left(\frac{4}{10}\right)^{3n-1} = \left(\frac{4}{10}\right)^{3n}\left(\frac{4}{10}\right)^{-1} = \frac{10}{4}\cdot\left(\left(\frac{4}{10}\right)^{3}\right)^{n}$$

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What law of exponents is used here, though? I understand that if it were $(4/10)^n$ I could factor a single $4/10$ out and then get $(4/10)(4/10)^{n-1}$, but I don't understand what happened here. –  agent154 May 27 '13 at 20:18
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This uses the property that for any $a, k, n \in \mathbb{R}$, we have $a^{kn} = (a^{k})^{n} = (a^{n})^{k}$ –  AWertheim May 27 '13 at 20:21
    
OK... I see it now. Thanks for the edit to clarify the steps. As I stated, I was aware of how to take one of the constants away, but not put one back. –  agent154 May 27 '13 at 20:24
    
Great! Glad I could help :) –  AWertheim May 27 '13 at 20:24

$$\sum_{n=1}^{\infty}\left(\frac{4}{10}\right)^{3n-1}=\frac{10}{4}\sum_{n=1}^{\infty}\left(\frac{64}{1000}\right)^{n}=\frac{10}{4}\times\frac{\frac{64}{1000}}{1-\frac{64}{1000}}$$

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