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

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

− | {{ | + | 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>. | ||

+ | |||

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

+ | |||

+ | <math>\sum_{k = 1}^{\infty}{\frac{8}{4^{k}}}</math>: | ||

+ | |||

+ | We can use the formula for the sum of an infinite geometric series (<math>\frac{a}{1-r}</math> where <math>a</math> is the first term and <math>r</math> is the common ratio). | ||

+ | <math>a=\frac{8}{4}=2</math> and <math>r=\frac{1}{4}</math> since each term is getting multiplied by <math>\frac{1}{4}</math> to receive the next term. Therefore, this sum is: | ||

+ | <math>\frac{2}{1-\frac{1}{4}}=\frac{2}{\frac{3}{4}}=\frac{8}{3}</math>. | ||

+ | |||

+ | <math>\sum_{k = 1}^{\infty}{\frac{2^k}{4^{k}}}</math>: | ||

+ | |||

+ | Similarly, we can use the formula used to solve the first part. | ||

+ | <math>a=\frac{2}{4}=\frac{1}{2}</math> and <math>r=\frac{1}{2}</math>. Therefore, this sum is: <math>\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1</math>. | ||

+ | |||

+ | Using these two answers, the desired sum is: <math>\frac{8}{3}+1=\boxed{\frac{11}{3}}</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: .