Trouble sum #1

Algebra Level 5

∑ ∑ ∑ ⏟ 1 ≤ a < b < c 1 2 a 3 b 5 c = 1 A \large \underbrace{\sum \sum\sum}_{1\le a<b<c} \frac{1}{2^a3^b5^c}=\frac{1}{A}

Find the value of A A .

Try part 2 here .


The answer is 1624.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Chew-Seong Cheong
Apr 16, 2018

S = ∑ ∑ ∑ ⏟ 1 ≤ a < b < c 1 2 a 3 b 5 c = ∑ a = 1 ∞ 1 2 a ∑ b = a + 1 ∞ 1 3 b ∑ c = b + 1 ∞ 1 5 c = ∑ a = 1 ∞ 1 2 a ∑ b = a + 1 ∞ 1 3 b ⋅ 1 5 b + 1 1 1 − 1 5 = ∑ a = 1 ∞ 1 2 a ∑ b = a + 1 ∞ 1 3 b ⋅ 1 5 b ⋅ 1 4 = 1 4 ∑ a = 1 ∞ 1 2 a ∑ b = a + 1 ∞ 1 1 5 b = 1 4 ∑ a = 1 ∞ 1 2 a 1 1 5 a + 1 ⋅ 15 14 = 1 56 ∑ a = 1 ∞ 1 3 0 a = 1 56 ⋅ 1 30 ⋅ 30 29 = 1 1624 \begin{aligned} S & = \underbrace{\sum \sum \sum}_{1\le a < b < c} \frac 1{2^a3^b5^c} \\ & = \sum_{a=1}^\infty \frac 1{2^a} \sum_{b=a+1}^\infty \frac 1{3^b} \sum_{c=b+1}^\infty \frac 1{5^c} \\ & = \sum_{a=1}^\infty \frac 1{2^a} \sum_{b=a+1}^\infty \frac 1{3^b} \cdot \frac 1{5^{b+1}} \frac 1{1-\frac 15} \\ & = \sum_{a=1}^\infty \frac 1{2^a} \sum_{b=a+1}^\infty \frac 1{3^b} \cdot \frac 1{5^b} \cdot \frac 14 \\ & = \frac 14 \sum_{a=1}^\infty \frac 1{2^a} \sum_{b=a+1}^\infty \frac 1{15^b} \\ & = \frac 14 \sum_{a=1}^\infty \frac 1{2^a} \frac 1{15^{a+1}} \cdot \frac {15}{14} \\ & = \frac 1{56} \sum_{a=1}^\infty \frac 1{30^a} \\ & = \frac 1{56} \cdot \frac 1{30} \cdot \frac {30}{29} \\ & = \frac 1{1624} \end{aligned}

Therefore, A = 1624 A = \boxed{1624} .

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