Difference between revisions of "G285 2021 Fall Problem Set Problem 8"
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==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} | + | 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>. |
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> | 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 ![\[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a+2b+3c}{4^{(a+b+c)}}\]](http://latex.artofproblemsolving.com/e/9/1/e91eecbe57dc0b8011bc0ad39d2b05b34ed54e32.png)
can be represented as 
, where 
and 
are relatively prime. Find 
.
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 $\frac{1}{4^{b+c}=\left(\frac{4}{3} \right)^2 = \frac{16}{9}$ (Error compiling LaTeX. Unknown error_msg), so the entire sum must be 
.
Now, coming back to the original problem, we split the single sum into 
: ![\[\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c)}}\]](http://latex.artofproblemsolving.com/0/d/9/0d9a338ced1a6932c6396bf5e0bfe855bc0227ac.png)
