jmc

Question

Let a and b be relatively prime positive integers such that ab=12^1+23^2+32^3+43^4+52^5+63^6+, where the numerators always increase by 1, and the denominators alternate between powers of 2 and 3, with exponents also increasing by 1 for each subsequent term. Compute a+b.

Accepted Answer

The sum can be split into two groups of numbers that we want to add: 12 + 32^3 + 52^5 and 23^2 + 43^4 + 63^6 Let X be the sum of the first sequence, so we havealign* X &= 12 + 32^3 + 52^5 X4 &= 0 + 12^3 + 32^5 34X &= 12 + 22^3 + 22^5 34X &= 12 + 1434 34X &= 56 X &= 109 align* Let Y be the sum of the second sequence, so we havealign* Y &= 23^2 + 43^4 + 63^6 19Y &= 0 + 23^4 + 43^6 89Y &= 23^2 + 23^4 + 23^6 89Y &= 2989 Y &= 14 98 &= 932 align*That means ab = 109 + 932 = 401288, so a+b = 689.