Question

在等比数列an中,若q=1/2,且a3+a6+a9+…+a99=60,则a1+a2+a3+…+a99=?

Answer 1

a3+a6+a9+…+a99 构成了公比为 (1/2)^3 = 1/8 的等比数列,共 (99-3)/3+1 = 33 项

a3 = a1*(1/2)^2 = a1 / 4

a3 * (1-(1/8)^33) / (1-1/8) = a1 /4 * (1-(1/8)^33) / (1-1/8) = 60

a1+a2+a3+…+a99 = a1(1-(1/2)^99) / (1-1/2) = 60 * 7/8 * 4 * 2 = 420