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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 $r$.

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

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  • $\begingroup$ What? You have that $$\sum\limits_{n=0}^\infty \frac{2^n + 4^n}{6^n} = \sum\limits_{n=0}^\infty \left(\frac13\right)^n + \sum\limits_{n=0}^\infty \left(\frac23\right)^n$$ by the calculation in my answer. $\endgroup$ – Eff Nov 28 '16 at 2:40
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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.

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HINT:

Write this as the sum of two geometric series.

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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

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