已知数列{an}的前n项和Sn=n（bn）,其中{bn}是首项为1,公差为2的等差数列
（1）求an的通项公式；（2）若cn=1/[an(2bn+3)],求数列｛cn｝的前n项和Tn
人气：412 ℃ 时间：2019-10-29 22:01:01
解答
(1)bn=1+(n-1)*2=2n-1
Sn=n(2n-1)
当n>2orn=2时:an=Sn-Sn-1=4n-3
n=1时a1=S1=1
所以an=4n-3
(2)Cn=1/[(4n-3)(4n+1)]=(1/4)*{1/(4n-3)-1/(4n+1)}
Tn=(1/4)*[1-1/(4n+1)]
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