数列{an}通项公式为an=1n(n+2),则数列{an}前n项和为Sn=______.-数学

题目简介

数列{an}通项公式为an=1n(n+2),则数列{an}前n项和为Sn=______.-数学

题目详情

数列{an}通项公式为an=
1
n(n+2)
,则数列{an}前n项和为Sn=______.
题型:填空题难度:中档来源:不详

答案

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

Sn=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
n
-class="stub"1
n+2
)

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

故答案为class="stub"n+1
2(n+2)

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