数列12•5,15•8,18•11…1(3n-1)(3n+2),…的前n项和Sn为()A.n3n+2B.n6n+4C.3n6n+4D.n+1n+2-数学

题目简介

数列12•5,15•8,18•11…1(3n-1)(3n+2),…的前n项和Sn为()A.n3n+2B.n6n+4C.3n6n+4D.n+1n+2-数学

题目详情

数列
1
2•5
1
5•8
1
8•11
1
(3n-1)(3n+2)
,…的前n项和Sn为(  )
A.
n
3n+2
B.
n
6n+4
C.
3n
6n+4
D.
n+1
n+2
题型:单选题难度:中档来源:不详

答案

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

Sn=class="stub"1
2•5
+class="stub"1
5•8
+…+class="stub"1
(3n-1)(3n+2)

=class="stub"1
3
(class="stub"1
2
-class="stub"1
5
+class="stub"1
5
-class="stub"1
8
+…+class="stub"1
3n-1
-class="stub"1
3n+2
)

=class="stub"1
3
(class="stub"1
2
-class="stub"1
3n+2
)
=class="stub"n
6n+4

故选B

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