题目

设函数f(x)=logmx，数列{an}是公比为m的等比数列，若f(a2a4…a2 006)=8，则f(a12)+f(a22)+…+f(a2 0062)的值等于A.-1 974                B.-1 990               C.2 022                 D.2 038 答案：A  f(a2a4…a2 006)=8logm［(a1m)(a3m)(a5m)…(a2 005m)］=8logm［(a1a3a5…a2 005)·m1 003］=8logm(a1a3a5…a2 005)+logmm1 003=8logm(a1a3a5…a2 0005)=8-1 003.又f(a12)+f(a22)+…+f(a2 0062)=2logm(a1a2a3…a2 006)=2［logm(a1a3a5…a2 005)］+logm(a2a4…a2 006)=2［8+(8-1 003)］=-1 974.

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