Difference between revisions of "G285 2021 Fall Problem Set Problem 8"

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==Problem==
 
==Problem==
If the value of <cmath>\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{a+2b+3c}{4^{(a+b+c)}}</cmath> can be represented as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime. Find <math>m+n</math>.
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If the value of <cmath>\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c)}}</cmath> can be represented as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime. Find <math>m+n</math>.
  
 
==Solution==
 
==Solution==
Geometric series spam
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We begin with a simpler problem <cmath>\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{1}{4^{(a+b+c)}}</cmath>. Now, suppose <math>a</math> and <math>b</math> are constant. We have a converging geometric series for <math>c</math> with a sum of <math>\frac{1}{1-\frac{1}{4}}=\frac{4}{3}</math>. Now, make <math>b</math> everchanging. We have <math>\frac{1}{4^{b+c})=\left(\frac{4}{3} \right)^2 = \frac{16}{9}</math>, so the entire sum must be <math>\frac{64}{27}</math>.
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Now, coming back to the original problem, we split the single sum into <math>3</math>: <cmath>\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c)}}</cmath>

Revision as of 21:19, 11 July 2021

Problem

If the value of \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c)}}\] can be represented as $\frac{m}{n}$, where $m$ and $n$ are relatively prime. Find $m+n$.

Solution

We begin with a simpler problem \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{1}{4^{(a+b+c)}}\]. Now, suppose $a$ and $b$ are constant. We have a converging geometric series for $c$ with a sum of $\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$. Now, make $b$ everchanging. We have $\frac{1}{4^{b+c})=\left(\frac{4}{3} \right)^2 = \frac{16}{9}$ (Error compiling LaTeX. Unknown error_msg), so the entire sum must be $\frac{64}{27}$.

Now, coming back to the original problem, we split the single sum into $3$: \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c)}}\]