已知Sn是等比数列{an}的前n项和,S3,S9,S6成等差数列
a,求数列{an}的公比q,b,若a1=1,求数列{n*a(3n-2)}(n属于N)的前n项和Tn
人气:352 ℃ 时间:2020-05-29 13:30:52
解答
a) 当q=1是,S3=3a1,S9=9a1,S6=6a1,则18a1=9a1 a1=0(舍)q不等于1时,S3=a1(1-q^3)/(1-q),S9=a1(1-q^9)/(1-q),S6=a1(1-q^6)/(1-q),2S9=S3+S6 有2q^9=q^3+q^62q^6=1+q^3 解得q=1(舍),q=-(1/2)^(1/3)b) an=a1q^(n-1) a(...
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