已知数列an的前项和为Sn,a1=1,nSn+1-(n+1)Sn=n^2+cn,S1,S2/2,S3/3成等差数列.(1)求C的值.(2)求数列{an}的通项公式.(3)设:Tn=1/S1+1/S2+…+1/Sn.求证:Tn
人气:124 ℃ 时间:2020-06-13 07:24:57
解答
nSn+1-(n+1)Sn=n^2+cn
两边同除以n(n+1)
=>Sn+1/(n+1)-Sn/n=(n+c)/(n+1)
S1,S2/2,S3/3成等差数列
=>c=1
Sn/n=n
=>Sn=n^2
=>an=2n-1
Tn=1+1/2^2+…+1/n^2
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