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==
{{potd_solution}}
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We can split this summation, as shown:
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<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>.
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Now we must simply find each of the smaller sums, and add them.
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<math>\sum_{k = 1}^{\infty}{\frac{8}{4^{k}}}</math>:
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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).
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<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:
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<math>\frac{2}{1-\frac{1}{4}}=\frac{2}{\frac{3}{4}}=\frac{8}{3}</math>.
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<math>\sum_{k = 1}^{\infty}{\frac{2^k}{4^{k}}}</math>:
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Similarly, we can use the formula used to solve the first part.
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<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>.
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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: $\sum_{k = 1}^{\infty}{\frac{8+2^{k}}{4^{k}}} =  \sum_{k = 1}^{\infty}{\frac{8}{4^{k}}} +\sum_{k = 1}^{\infty}{\frac{2^k}{4^{k}}}$.

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: $\frac{8}{3}+1=\boxed{\frac{11}{3}}$.

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