已知数列{an}满足前n项和为Sn=n2+1数列{bn}满足bn= ,且前n项和为Tn,设Cn=T2n+1-Tn
人气:312 ℃ 时间:2019-09-29 01:48:10
解答
(1) ∵数列{an}满足前N项和sn=n平方+1 ∴Sn=n^2+1 S(n-1)=(n-1)^2+1 An=Sn-S(n-1) =n^2+1-[(n-1)^2+1] =2n-1 A1=S1=2 Bn=2/An +1=2/(2n-1)+1=(2n+1)/(2n-1) B1=2/A1+1=2 Bn是一个首项为2,通项为(2n+1)/(2n-1) 的数列...
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