来自刘巧歌的问题
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.
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.
1回答
2020-12-19 22:12
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.
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.
a7=a1q^6a4=a1q^3因为a1,a7,a4成等差数列所以a7=(a1a4)/2a1q^6=(a1a1q^3)/22q^6-q^3-1=0(2q^31)(q^3-1)=0q^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=(1q^3)/2(S12-S6)/S6=(1q^6)-1当q^3=1时,S6/2S3=(1q^3)/2=1,(S12-S6)/S6=(1q^6)-1=1当q^3=-1/2时,S6/2S3=(1q^3)/2=1/4,(S12-S6)/S6=(1q^6)-1=1/4所以2S3,S6,S12-S6成等比数列