设{an}是公比大于1的等比数列,Sn为数列{an}的前n项和,已知S3=7,且a1+3,2a2,a3+4
构成等差数列
1.求数列{an}的通项
2.令 bn=ln [ a(3n+1) ] ,n=1,2,…,求数列{bn}的前n项和Tn
人气:124 ℃ 时间:2019-12-20 15:41:36
解答
1.a1+3,2a2,a3+4构成等差数列
可得4a2=a1+a3+7
又a3=a1q^2,a2=a1q,故4a1q=a1+a1q^2+7,
又S3=7,解得a1=1,q=2
则an=2^(n-1)
2. bn=ln [ a(3n+1)]= bn=ln [ 2^(3n)]=3nln2
故Tn=(3+6+……+3n)ln2=1/2(3n+3n^2)ln2
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