设数列{xn}满足logaxn+1＝1+logaxn，且x1+x2+…+x100=100，x101+x102+…+x200的值为（）A.100aB.101a2C.101a100D.100a100

Question

设数列{ x_n}满足logaxn+1＝1+logaxn，且x₁+x₂+…+x₁₀₀=100，x₁₀₁+x₁₀₂+…+x₂₀₀的值为（　　）
A. 100a
B. 101a²
C. 101a¹⁰⁰
D. 100a¹⁰⁰

Answer 1

∵log_ax_n+1=1+log_ax_n，∴log_ax_n+1-log_ax_n=1，
∴

log	(xn+1)/(xn) a

=1，则(xn+1)/(xn)=a，
∴数列{x_n}是以a为公比的等比数列，
∵x₁+x₂+…+x₁₀₀=100，∴x₁₀₁+x₁₀₂+…+x₂₀₀=a¹⁰⁰x₁+a¹⁰⁰x₂+…a¹⁰⁰x₁₀₀
=a¹⁰⁰（x₁+x₂+…+x₁₀₀）=100a¹⁰⁰，
故选D．