已知数列1/6,1/12,1/20,…,1/(n+1)(n+2)…,则其前n项和Sn=_.

问题描述:

已知数列

1
6
1
12
1
20
,…,
1
(n+1)(n+2)
…,则其前n项和Sn=______.

设数列为{an}则由题意可得:
数列的通项公式为an =

1
(n+1)(n+2)
1
n+1
1
n+2

所以Sn=a1+a2+…+an
=
1
2
1
3
+
1
3
1
4
+… +
1
n+1
1
n+2

=
1
2
1
n+2
2
2(n+2)

故答案为
2
2(n+2)