已知数列16,112,120,…,1(n+1)(n+2)…,则其前n项和Sn=______.-数学

题目简介

已知数列16,112,120,…,1(n+1)(n+2)…,则其前n项和Sn=______.-数学

题目详情

已知数列
1
6
1
12
1
20
,…,
1
(n+1)(n+2)
,则其前n项和Sn=______.
题型:填空题难度:中档来源:不详

答案

设数列为{an}则由题意可得:
数列的通项公式为an =class="stub"1
(n+1)(n+2)
=class="stub"1
n+1
-class="stub"1
n+2

所以Sn=a1+a2+…+an
=class="stub"1
2
-class="stub"1
3
+class="stub"1
3
-class="stub"1
4
+… +class="stub"1
n+1
-class="stub"1
n+2

=class="stub"1
2
-class="stub"1
n+2
=class="stub"2
2(n+2)

故答案为class="stub"2
2(n+2)

更多内容推荐