Finding the sum of series $\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}$

Question

I am being asked to find the sum of the following convergent series :

$$\sum_{n=0}^∞ \frac{2^n}{6^n} + \frac{3^n}{6^n}$$

Attempting to generalize from partial sums yields nothing of interest:

$s_1 = \frac{5}{6}$

$s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43}{36}$

$s_3 = \frac{43}{36} + \frac{35}{216} = \frac{293}{216}$

$s_4 = \frac{293}{216} + \frac{97}{1296} = \frac{1855}{1296} $

I do not see a pattern here...

How must I proceed to find the sum of this series?

Answer 1

It is just the sum of two geometric series in disguise.

$$ \begin{aligned} \underset{i=0}{\overset{\infty}{\sum}}\frac{2^{n}+3^{n}}{6^{n}}&=\underset{i=0}{\overset{\infty}{\sum}}\frac{2^{n}}{6^{n}}+\frac{3^{n}}{6^{n}}\\ &=\underset{i=0}{\overset{\infty}{\sum}}(1/3)^{n}+\underset{i=0}{\overset{\infty}{\sum}}(1/2)^{n}\\ &=\frac{1}{1-\frac{1}{3}}+\frac{1}{1-\frac{1}{2}}\\ &=\frac{3}{2}+2\\ &=\frac{7}{2}. \end{aligned} $$

Answer 2

$\frac{2^n}{6^n}=(\frac{1}{3})^n$ and $\frac{3^n}{6^n}=(\frac{1}{2})^n$

Now, use the well known result that for $|x|<1$,

$$\sum \limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$$

Answer 3

Note that the sum $\sum_{n=0}^{\infty}r^{n}=\frac{1}{1-r}$ for $|r|<1$. According to this fact $\sum_{n=0}^{\infty}\frac{2^{n}+3^{n}}{6^{n}}=\sum_{n=0}^{\infty}\frac{2^{n}}{6^{n}}+\sum_{n=0}^{\infty}\frac{3^{n}}{6^{n}}=\sum_{n=0}^{\infty}(\frac{2}{6})^{n}+\sum_{n=0}^{\infty}(\frac{3}{6})^{n}=\frac{1}{1-1/3}+\frac{1}{1-1/2}=\frac{7}{2}$.