Question

设{an}是公比不等于1的等比数列,Sn是其前n项和,若a1,2a7,3a4成等差数列,求证12S3,S6,S12-S6

Answer 1

q=公比≠1

a7=a4q^3

4a7=a1+3a4=4a4q^3

a4=a1/(4q^3-3)=a1q^3,Q=q^3

1=(4Q-3)Q=4Q^2-3Q

4Q^2-3Q-1=0

(4Q+1)(Q-1)=0

Q=-1/4

q=(-1/4)^(1/3)

令z=1-q

12S3=12a1(1-q^3)/(1-q)=15a1/(1-q)=15a1/z

S6=a1(1-1/16)/(1-q)=15a1/16z

S12-S6=(a1/z)[1-1/16^2-1+1/16]

=(a1/z)*15/16^2

S6/12S3=[(a1/z)*15/16]/15(a1/z)=1/16

(S12-S6)S6=(15/16^2)/（15/16）=1/16=S6/12S3

所以,12S3,S6,S12-S6成等比数列