G285 2021 Fall Problem Set Problem 8

Problem

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

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

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 $3$ : $\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{a}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{2b}{4^{(a+b+c)}}+\sum_{a=0}^{\infty} \sum_{b=0}^{\infty} \sum_{c=0}^{\infty} \frac{3c}{4^{(a+b+c)}}$ 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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G285 2021 Fall Problem Set Problem 8

Problem

Solution