Question

Сократите дробь
(18^(n+1))/((3^(2n+5))*(2^(n-2)))

Answer 1

(18^(n+1))/((3^(2n+5))*(2^(n-2)))=(2^(n+1))(9^(n+1))/((3^(2n+5))*(2^(n-2)))=(2^(n+1))(3^(2n+2))/((3^(2n+5))*(2^(n-2)))=(2^(n+1-n+2))(3^(2n+2-2n-5))=2^3*3^-3=8/27