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