Question

设函数f(x)=log_mx，数列{a_n}是公比为m的等比数列，若f(a₂a₄…a_{2 006})=8，则f(a₁²)+f(a₂²)+…+f(a_{2 006}²)的值等于

A.-1 974                B.-1 990               C.2 022                 D.2 038

Answer 1

A 

解析：f(a₂a₄…a_{2 006})=8log_m［(a₁m)(a₃m)(a₅m)…(a_{2 005}m)］=8log_m［(a₁a₃a₅…a_{2 005})·m^{1 003}］=8log_m(a₁a₃a₅…a_{2 005})+log_mm^{1 003}=8log_m(a₁a₃a₅…a_{2 0005})=8-1 003.

又f(a₁²)+f(a₂²)+…+f(a_{2 006}²)=2log_m(a₁a₂a₃…a_{2 006})

=2［log_m(a₁a₃a₅…a_{2 005})］+log_m(a₂a₄…a_{2 006})

=2［8+(8-1 003)］=-1 974.