数列11+2,11+2+3,…11+2+…+n的前n项和为()A.nn+1B.2nn+1C.nn+2D.n2(n+1)-高二数学

题目简介

数列11+2,11+2+3,…11+2+…+n的前n项和为()A.nn+1B.2nn+1C.nn+2D.n2(n+1)-高二数学

题目详情

数列
1
1+2
1
1+2+3
,…
1
1+2+…+n
的前n项和为(  )
A.
n
n+1
B.
2n
n+1
C.
n
n+2
D.
n
2(n+1)
题型:单选题难度:偏易来源:不详

答案

由数列可知数列的通项公式an=class="stub"1
1+2+…+(n+1)
=class="stub"1
(n+1)(n+2)
2
=class="stub"2
(n+1)(n+2)
=2(class="stub"1
n+1
-class="stub"1
n+2
)

∴数列的前n项和S=2(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
)=2(class="stub"1
2
-class="stub"1
n+2
)=class="stub"n
n+2

故选:C.

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