Compute \(\sum_{1 \le a < b < c} \frac{1}{2^a 3^b 5^c}\)
(The sum is taken over all triples \((a,b,c)\) of positive integers such that \(1 \le a < b < c.\))
+6248 \(\sum \limits_{a=1}^\infty~\sum \limits_{b=a+1}^\infty~\sum \limits_{c=b+1}^\infty~\dfrac{1}{2^a3^b5^c} = \\ \sum \limits_{a=1}^\infty~\sum \limits_{b=a+1}^\infty~\dfrac{1}{2^a3^b}\sum \limits_{c=b+1}^\infty~\dfrac{1}{5^c}=\\ \sum \limits_{a=1}^\infty~\sum \limits_{b=a+1}^\infty~\dfrac{1}{2^a3^b}\sum \limits_{c=0}^\infty~\dfrac{1}{5^{c+b+1}}=\\ \sum \limits_{a=1}^\infty~\sum \limits_{b=a+1}^\infty~\dfrac{1}{2^a3^b}\dfrac{1}{5^{b+1}}\dfrac{5}{4} =\)
\(\sum \limits_{a=1}^\infty~\sum \limits_{b=a+1}^\infty~\dfrac{1}{2^a3^b}\dfrac{1}{5^{b+1}}\dfrac{5}{4} =\\ \sum \limits_{a=1}^\infty~\dfrac{1}{2^a}~\sum \limits_{b=a+1}^\infty~\dfrac{1}{15^b}\dfrac 1 5\dfrac 5 4=\\ \dfrac 1 4 \sum \limits_{a=1}^\infty~\dfrac{1}{2^a}~\sum \limits_{b=0}^\infty~\dfrac{1}{15^{b+a+1}}=\\ \dfrac 1 4 \sum \limits_{a=1}^\infty~\dfrac{1}{2^a}\dfrac{1}{15^{a+1}}\dfrac{15}{14} = \\ \dfrac 1 4 \sum \limits_{a=1}^\infty~\dfrac{1}{30^a}\dfrac{1}{15}\dfrac{15}{14} = \\ \dfrac{1}{56}\sum \limits_{a=1}^\infty~\dfrac{1}{30^a} =\)
\(\dfrac{1}{56}\dfrac{1}{30}\dfrac{30}{29} = \dfrac{1}{56}\dfrac{1}{29} = \dfrac{1}{1624}\)
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