Putnam 2015, Problem B-4

Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a,b,c$. Express
$$\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c}$$
as a rational number in lowest terms.

Solution:
First solution: The answer is \(17/21\). For fixed \(b, c\), there is a triangle of side lengths \(a, b, c\) if and only if \(|b - c| < a < b + c\). It follows that the desired sum is
\[
S = \sum_{b,c} \frac{1}{3^b 5^c} \left( \sum_{a=|b-c|+1}^{b+c-1} 2^a \right) = \sum_{b,c} \frac{2^{b+c} - 2^{|b-c|+1}}{3^b 5^c}.
\]
We write this as \(S = S_1 + S_2\) where \(S_1\) sums over positive integers \(b, c\) with \(b \leq c\) and \(S_2\) sums over \(b > c\). Then
\[
\begin{align*}
S_1 &= \sum_{b=1}^\infty \sum_{c=b}^\infty \frac{2^{b+c} - 2^{c-b+1}}{3^b 5^c} \\
&= \sum_{b=1}^\infty \left( \left( \frac{2}{3} \right)^b - \frac{2}{6^b} \right) \sum_{c=b}^\infty \left( \frac{2}{5} \right)^c \\
&= \sum_{b=1}^\infty \left( \left( \frac{2}{3} \right)^b - \frac{2}{6^b} \right) \frac{5}{3} \left( \frac{2}{5} \right)^b \\
&= \sum_{b=1}^\infty \left( \frac{5}{3} \left( \frac{4}{15} \right)^b - \frac{10}{3} \left( \frac{1}{15} \right)^b \right) \\
&= \frac{85}{231}.
\end{align*}
\]