# Difference between revisions of "AoPS Wiki talk:Problem of the Day/August 1, 2011"

## Solution

We can split this summation, as shown: $\sum_{k = 1}^{\infty}{\frac{8+2^{k}}{4^{k}}} = \sum_{k = 1}^{\infty}{\frac{8}{4^{k}}} +\sum_{k = 1}^{\infty}{\frac{2^k}{4^{k}}}$.

Now we must simply find each of the smaller sums, and add them. $\sum_{k = 1}^{\infty}{\frac{8}{4^{k}}}$:

We can use the formula for the sum of an infinite geometric series ( $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio). $a=\frac{8}{4}=2$ and $r=\frac{1}{4}$ since each term is getting multiplied by $\frac{1}{4}$ to receive the next term. Therefore, this sum is: $\frac{2}{1-\frac{1}{4}}=\frac{2}{\frac{3}{4}}=\frac{8}{3}$. $\sum_{k = 1}^{\infty}{\frac{2^k}{4^{k}}}$:

Similarly, we can use the formula used to solve the first part. $a=\frac{2}{4}=\frac{1}{2}$ and $r=\frac{1}{2}$. Therefore, this sum is: $\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$.

Using these two answers, the desired sum is: $\frac{8}{3}+1=\boxed{\frac{11}{3}}$.