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在等比数列{an}中,a1+a3=30,a5+a7=120,求a9+a11=?
人气:458 ℃ 时间:2020-02-27 16:50:47
解答
an = a1q^(n-1)
a1+a3=30
a1(1+q^2) =30(1)
a5+a7=120
a1q^4(1+q^2)=120(2)
(2)/(1)
q^4= 40
a9+a11
=a1q^8(1+q^2)
= 30q^8( a1(1+q^2) =30 )
= 30(40)^2 (q^4=40)
=48000
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