ВП
Владимир Пронин
Ответ
[3^(n) + 3^(n-1)]^2 / 9^(n-1) =
= [3^(n) + 3^(n-1)]^2 / [(3^2)^(n-1) =
= [3^(n) + 3^(n-1)]^2 / [3^{2(n-1)}] =
= [ {3^(n) + 3^(n-1)} / 3^(n-1) ]^2 =
= [ 3^(n) / 3^(n-1) + 3^(n-1) / 3^(n-1) ]^2 =
= [3^{n-(n-1)} + 3^{(n-1)-(n-1)}]^2 =
= [3^(1) + 1]^2 = 4^2 = 16