Difference between revisions of "G285 2021 Fall Problem Set Problem 8"
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==Problem== | ==Problem== | ||
| − | + | 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> | |
| + | |||
| + | <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== | ||
| − | + | 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+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 ![\[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \sum_{d=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c+d)}}\]](http://latex.artofproblemsolving.com/4/7/1/4716a882e6c5ff7f1d14bb7983f03c8128fc35b3.png)
![\[\textbf{(A)}\ \frac{16}{27} \qquad \textbf{(B)}\ \frac{32}{27} \qquad \textbf{(C)}\ \frac{64}{27} \qquad \textbf{(D)}\ \frac{128}{27} \qquad \textbf{(E)}\ \frac{256}{27}\]](http://latex.artofproblemsolving.com/a/f/d/afdbf5d39489af9cd30f524bc1c1fe468a088206.png)
Solution
We begin with a simpler problem ![\[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{1}{4^{(a+b+c)}}\]](http://latex.artofproblemsolving.com/0/e/0/0e01ec19f235bbece9208e5c66e1ca8dc3963a54.png)
. Now, suppose 
and 
are constant. We have a converging geometric series for 
with a sum of 
. Now, make everchanging. We have 
, so the entire sum must be 
.
Now, coming back to the original problem, we split the single sum into 
: ![\[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a}{4^{(a+b+c+d)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c+d)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c+d)}}\]](http://latex.artofproblemsolving.com/8/1/5/81501e4dece6413465f03ecde2b3fe57a26e19bf.png)
Split into single variables to get ![\[\frac{16}{9} \left(\sum_{a=0}^{\infty} \frac{a}{4^a}+2\sum_{b=0}^{\infty} \frac{b}{4^b} + 3\sum_{c=0}^{\infty} \frac{c}{4^c} \right)\]](http://latex.artofproblemsolving.com/a/d/1/ad181aad6144dadac751e24c8f25a5ea0d95aa0c.png)
Now, generalize 
to obtain 
. Using the geometric series formula we have ![\[\frac{(\tfrac{1}{4}+\tfrac{1}{16}+\tfrac{1}{64}+ \cdots)}{1-\frac{1}{4}} \implies \frac{\tfrac{\tfrac{1}{4}}{1-\frac{1}{4}}}{1-\frac{1}{4}} \implies \frac{4}{9}\]](http://latex.artofproblemsolving.com/7/b/3/7b3d17051ff00a86b51172b2ae6b54e7193fa91f.png)
Now, we can plug this in for all 
to get ![\[\frac{16}{9} \left(6 \cdot \frac{4}{9} \right) \implies \boxed{\textbf{(D)}\ \frac{128}{27}}\]](http://latex.artofproblemsolving.com/e/b/b/ebb0fb2b16f30a936c77c882a0eb5a2d4ae36dcd.png)
