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等比数列a1+a2=3,a3+a4=6,求S8
人气:202 ℃ 时间:2020-03-18 04:16:54
解答
a3+a4=a1q^2+a2q^2=q^2(a1+a2)
6=3q^2
q^2=2
a5+a6
=q^2(a3+a4)
=2*6
=12
a7+a8
=q^2(a5+a6)
=2*12
=24
s8=a1+a2+a3+a4+a5+a6+a7+a8
=3+6+12+24
=45
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