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Algebra Difficulty 5.2 AIME, harder Find the answer

Find the value of a=1b=1c=1ab(3a+c)4a+b+c(a+b)(b+c)(c+a)\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{a b(3 a+c)}{4^{a+b+c}(a+b)(b+c)(c+a)}

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let SS denote the given sum. By summing over all six permutations of the variables a,b,ca, b, c we obtain 6S=a=1b=1c=13(a2b+a2c+b2a+b2c+c2a+c2b)+6abc4a+b+c(a+b)(b+c)(c+a)=a=1b=1c=134a+b+c=3(a=114a)(b=114b)(c=114c)=3(13)3=19\begin{aligned} 6 S & =\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{3\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b\right)+6 a b c}{4^{a+b+c}(a+b)(b+c)(c+a)} \\ & =\sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \sum_{c=1}^{\infty} \frac{3}{4^{a+b+c}} \\ & =3\left(\sum_{a=1}^{\infty} \frac{1}{4^{a}}\right)\left(\sum_{b=1}^{\infty} \frac{1}{4^{b}}\right)\left(\sum_{c=1}^{\infty} \frac{1}{4^{c}}\right) \\ & =3\left(\frac{1}{3}\right)^{3} \\ & =\frac{1}{9} \end{aligned} Hence S=154S=\frac{1}{54}.

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