已知数列{an}是首项a1=1/4,公比q=1/4的等比数列,设(bn)+2=3log1/4an(n∈N*),数列{cn}=an+bn
求证(1)bn是等差数列
(2)数列cn的前n项和Sn
人气:405 ℃ 时间:2020-02-01 07:09:01
解答
an= (1/4)^n
bn + 2 = 3log an
= 3n
bn = 3n-2
=>bn是等差数列
cn = an+bn
= (1/4)^n + 3n-2
Sn = c1+c2+..+cn
= (1/4)(1- (1/4)^n) /(1-1/4) + 3n(n+1)/2 - 2n
= (1/3)(1-(1/4)^n) + 3n(n+1)/2 - 2n
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