Basic Infinite Series Question: $\sum_{n=0}^\infty\frac{2^n+4^n}{6^n}=?$

Question

The question:

$$\sum_{n=0}^\infty\frac{2^n+4^n}{6^n}=\;?$$

I can't figure out how to approach this question as I can't find a constant value for $r$.

Answer 1

Split up the summand into $(\frac{2}{6})^n + (\frac{4}{6})^n = (\frac{1}{3})^n + (\frac{2}{3})^n$

That's the sum of two geometric series.

Answer 2

HINT:

Write this as the sum of two geometric series.

Answer 3

Hint. $$\frac{2^n+4^n}{6^n} = \frac{2^n}{6^n}+\frac{4^n}{6^n} = \left(\frac{2}{6}\right)^n+\left(\frac{4}{6}\right)^n = \left(\frac{1}{3}\right)^n+\left(\frac{2}{3}\right)^n.$$

Answer 4

Use this law: $$ \sum a+b = \sum a +\sum b $$

You get $$ \sum \frac{2^{n}+4^{n}}{6^{n}} = \sum \frac{2^{n}}{6^{n}} +\sum \frac{4^{n}}{6^{n}} $$ $$=\sum (\frac{2}{6})^{n} + \sum (\frac{4}{6})^{n}$$

where your two r values are 2/6 and 4/6