数列{an}中,an=1n(n+1)(n+2),Sn为{an}的前n项和,则S1+S2+…+S10的值为()A.5524B.124C.552D.6524-数学

题目简介

数列{an}中,an=1n(n+1)(n+2),Sn为{an}的前n项和,则S1+S2+…+S10的值为()A.5524B.124C.552D.6524-数学

题目详情

数列{an}中,an=
1
n(n+1)(n+2)
,Sn为{an}的前n项和,则S1+S2+…+S10的值为(  )
A.
55
24
B.
1
24
C.
55
2
D.
65
24
题型:单选题难度:中档来源:不详

答案

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

Sn=class="stub"1
2
[1-class="stub"1
n+1
-(class="stub"1
2
-class="stub"1
n+2
)]
=class="stub"1
4
+class="stub"1
2
(class="stub"1
n+2
-class="stub"1
n+1
)

∴S1+S2+…+S10=class="stub"10
4
+class="stub"1
2
[(class="stub"1
3
-class="stub"1
2
)+(class="stub"1
4
-class="stub"1
3
)+(class="stub"1
5
-class="stub"1
4
)+…+(class="stub"1
12
-class="stub"1
11
)]

=class="stub"5
2
+class="stub"1
2
(class="stub"1
12
-class="stub"1
2
)

=class="stub"55
24

故选A.

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