已知等比数列{an}中an>0,a1+a2+...+a8=4,a1*a2*...*a8=16,则1/a1+1/a2+...+1/a8的值为A.2B.4C.8D.16

Question

已知等比数列{an}中an>0,a1+a2+...+a8=4,a1*a2*...*a8=16,则1/a1+1/a2+...+1/a8的值为 A.2 B.4 C.8 D.16

Answer 1

设公比为q
a1+a2+...+a8=a1*(q^8-1)/(q-1)=4 (1)
a1*a2*...*a8=a1^8*q^(1+2+...+7)=a1^8*q^28=16 a1^2*q^7=2 (2)
(1)÷(2) (1/a1)*(q-1/q^7)/(q-1)=2
所以1/a1+1/a2+...+1/a8=1/a1+(1/a1)*1/q+(1/a1)*(1/q)^2+...+(1/a1)*(1/q)^7
=(1/a1)*(1-1/q^8)/(1-1/q)
=(1/a1)(q-1/q^7)/(q-1)
=2
选A