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

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{{:AoPSWiki:Problem of the Day/August 1, 2011}} | {{:AoPSWiki:Problem of the Day/August 1, 2011}} | ||

==Solution== | ==Solution== | ||

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We can split this summation, as shown: | We can split this summation, as shown: | ||

<math> \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}}} </math>. | <math> \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}}} </math>. |

## Latest revision as of 14:20, 1 August 2011

## Problem

AoPSWiki:Problem of the Day/August 1, 2011

## Solution

We can split this summation, as shown: .

Now we must simply find each of the smaller sums, and add them.

:

We can use the formula for the sum of an infinite geometric series ( where is the first term and is the common ratio). and since each term is getting multiplied by to receive the next term. Therefore, this sum is: .

:

Similarly, we can use the formula used to solve the first part. and . Therefore, this sum is: .

Using these two answers, the desired sum is: .