好吧问答库 > 已知数列{an}的前n项和为Sn，且Sn=n（n+1）（n∈N*），（Ⅰ）求数列{an}的通项公式an（Ⅱ）数列{bn}的通

已知数列{an}的前n项和为Sn，且Sn=n（n+1）（n∈N*），（Ⅰ）求数列{an}的通项公式an（Ⅱ）数列{bn}的通

2025年04月07日 14:38
有1个网友回答
网友（1）：

（Ⅰ）∵数列{an}的前n项和为Sn，
且Sn=n（n+1）（n∈N*），
∴a1=S1=1×（1+1）=2，
an=Sn-Sn-1=n（n+1）-（n-1）n
=（n2+n）-（n2-n）
=2n．
（Ⅱ）∵an=2n，∴bn=

1
an?an+1
=
1
2n?2(n+1)
=
1
4
(
1
n
?
1
n+1
)，
∴Tn=
1
4
（1?
1
2
+
1
2
?
1
3
+…+
1
n
?
1
n+1
）
=
1
4
(1?
1
n+1
)
=
n
4n+4
．

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