Question

Сократить дробь: (202^2-108^2)*5^3n-5/125^n-1

Answer 1

$\frac{(202^2-108^2)*5^{3n-5}}{125^{n-1}}=\frac{(202-108)(202+108)*5^{3n-5}}{(5^3)^{n-1}}=\frac{94*310*5^{3n-5}}{5^{3n-3}}= \frac{94*310}{5^{3n-3-3n+5}} \\ = \frac{94*310}{5^2}=\frac{94*310}{25}=1165.6$