已知数列{an}是等比数列,Sn是其前几项的和,a1,a7,a4成等差数列,
求证:2S1,S6,S12-S6成等比数列.
人气:493 ℃ 时间:2020-03-26 05:55:53
解答
a7=a1q^6 a4=a1q^3 因为a1,a7,a4成等差数列 所以a7=(a1 a4)/2 a1q^6=(a1 a1q^3)/2 2q^6-q^3-1=0 (2q^3 1)(q^3-1)=0 q^3=1,或q^3=-1/2 因为 2S3=2a1(1-q^3)/(1-q) S6=a1(1-q^6)/(1-q) S12-S6=a1(1-q^12)/(1-q)-a1(1-q^6)/(1-q) S6/2S3=(1 q^3)/2 (S12-S6)/S6=(1 q^6)-1 当q^3=1时,S6/2S3=(1 q^3)/2=1,(S12-S6)/S6=(1 q^6)-1 =1 当q^3=-1/2时,S6/2S3=(1 q^3)/2=1/4,(S12-S6)/S6=(1 q^6)-1 =1/4 所以2S3,S6,S12-S6成等比数列
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