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

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==Problem==
 
==Problem==
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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>.
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Find <cmath>\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \sum_{d=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c+d)}}</cmath>
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<cmath>\textbf{(A)}\ \frac{16}{27} \qquad \textbf{(B)}\ \frac{32}{27} \qquad \textbf{(C)}\ \frac{64}{27} \qquad \textbf{(D)}\ \frac{128}{27} \qquad \textbf{(E)}\ \frac{256}{27}</cmath>
  
 
==Solution==
 
==Solution==
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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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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>
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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+d)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c+d)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c+d)}}</cmath> Split into single variables to get <cmath>\frac{16}{9} \left(\sum_{a=0}^{\infty} \frac{a}{4^a}+2\sum_{b=0}^{\infty} \frac{b}{4^b} + 3\sum_{c=0}^{\infty} \frac{c}{4^c} \right)</cmath> Now, generalize <math>\sum_{x=0} \frac{x}{4^x}</math> to obtain <math>(\frac{1}{4}+\frac{1}{16}+ \frac{64}+ \cdots )+(\frac{1}{16}+\frac{64}+ \cdots)+(\frac{1}{64}+ \cdots )+ \cdots</math>. Using the geometric series formula we have <cmath>\frac{(\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+ \cdots)}{1-\frac{1}{4}} \implies \frac{\tfrac{\tfrac{1}{4}}{1-\frac{1}{4}}}{1-\frac{1}{4}} \implies \frac{4}{9}</cmath> Now, we can plug this in for all <math>(a,b,c)</math> to get <cmath>\frac{16}{9} \left(6 \cdot \frac{4}{9} \right) \implies \boxed{\textbf{(D)}\ \frac{128}{27}}</cmath>

Latest revision as of 20:44, 18 July 2021

Problem

Find \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \sum_{d=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c+d)}}\]

\[\textbf{(A)}\ \frac{16}{27} \qquad \textbf{(B)}\ \frac{32}{27} \qquad \textbf{(C)}\ \frac{64}{27} \qquad \textbf{(D)}\ \frac{128}{27} \qquad \textbf{(E)}\ \frac{256}{27}\]

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 $\frac{1}{4^{b+c}}=\left(\frac{4}{3} \right)^2 = \frac{16}{9}$, so the entire sum must be $\frac{64}{27}$.

Now, coming back to the original problem, we split the single sum into $3$: \[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a}{4^{(a+b+c+d)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c+d)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c+d)}}\] Split into single variables to get \[\frac{16}{9} \left(\sum_{a=0}^{\infty} \frac{a}{4^a}+2\sum_{b=0}^{\infty} \frac{b}{4^b} + 3\sum_{c=0}^{\infty} \frac{c}{4^c} \right)\] Now, generalize $\sum_{x=0} \frac{x}{4^x}$ to obtain $(\frac{1}{4}+\frac{1}{16}+ \frac{64}+ \cdots )+(\frac{1}{16}+\frac{64}+ \cdots)+(\frac{1}{64}+ \cdots )+ \cdots$. Using the geometric series formula we have \[\frac{(\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+ \cdots)}{1-\frac{1}{4}} \implies \frac{\tfrac{\tfrac{1}{4}}{1-\frac{1}{4}}}{1-\frac{1}{4}} \implies \frac{4}{9}\] Now, we can plug this in for all $(a,b,c)$ to get \[\frac{16}{9} \left(6 \cdot \frac{4}{9} \right) \implies \boxed{\textbf{(D)}\ \frac{128}{27}}\]

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