Question

В геометрической прогрессии (bn): b1=1/9; b7=81. Найдите (b4)^2+b3

Answer 1

bn = b1 * q^(n-1)
q^6 = b7/b1 = 81/ 1/9 = 81 * 9 = 729
q = 3
b3 = b1 * q^2 = 1/9 * 3^2 = 1/9 * 9 = 1
b4 = b1 * q^3 = 1/9 * 3^3 = 1/9 * 27 = 3
(b4)^2+b3 = 3^2 + 1 = 9 + 1 = 10

Answer 2

спасибоо))