S[n]是等比数列{a[n]}前n项和,若S[1],2S[2],3S[3]成等差数列,求证:{a[n]}公比是1/3
人气:412 ℃ 时间:2020-05-12 04:21:49
解答
S1=A1
S2=A1+A2=A1*(1+q)
S3=A1+A2+A3=A1*(1+q+q^2)
所以:A1+3A1*(1+q+q^2)=4A1*(1+q)
因为{An}是等比数列,所以A1不等于0
两边同时约去A1,所以q=0(舍去)或者q=1/3
S1,2S2,3S3成等差数列
4S2=S1+3S3
4(a1+a2)=a1+3*(a1+a2+a3)
4a1+4a2=a1+3a1+3a2+3a3
a2=3a3
q=a3/a2=1/3
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