View source for G285 2021 Fall Problem Set Problem 8

==Problem==
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==
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>