jmc

Question

Let _n = 0^123456789 3n^2 + 9n + 7(n^2 + 3n + 2)^3 = ab,where a and b are relatively prime positive integers. Find b - a.

Accepted Answer

We can write align* 3n^2 + 9n + 7(n^2 + 3n + 2)^3 &= 3n^2 + 9n + 7(n + 1)^3 (n + 2)^3 &= (n^3 + 6n^2 + 12n + 8) - (n^3 + 3n^2 + 3n + 1)(n + 1)^3 (n + 2)^3 &= (n + 2)^3 - (n + 1)^3(n + 1)^3 (n + 2)^3 &= 1(n + 1)^3 - 1(n + 2)^3. align*Therefore, align* _n = 0^123456789 3n^2 + 9n + 7(n^2 + 3n + 2)^3 &= _n = 0^123456789 ( 1(n + 1)^3 - 1(n + 2)^3 ) &= ( 1 - 12^3 ) + ( 12^3 - 13^3 ) + ( 13^3 - 14^3 ) + + ( 1123456790^3 - 1123456791^3 ) &= 1 - 1123456791^3 &= 123456791^3 - 1123456791^3. align*Thus, a = 123456791^3 - 1 and b = 123456791^3, so b - a = 1.