已知数列{an}:12,13+23,14+24+34,15+25+35+45,…,那么数列{bn}={1anan+1}前n项的和为()A.4(1-1n+1)B.4(12-1n+1)C.1-1n+1D.

题目简介

已知数列{an}:12,13+23,14+24+34,15+25+35+45,…,那么数列{bn}={1anan+1}前n项的和为()A.4(1-1n+1)B.4(12-1n+1)C.1-1n+1D.

题目详情

已知数列{an}:
1
2
1
3
+
2
3
1
4
+
2
4
+
3
4
1
5
+
2
5
+
3
5
+
4
5
,…
,那么数列{bn}={
1
anan+1
}
前n项的和为(  )
A.4(1-
1
n+1
)
B.4(
1
2
-
1
n+1
)
C.1-
1
n+1
D.
1
2
-
1
n+1
题型:单选题难度:偏易来源:不详

答案

数列{an}的通项公式为an=class="stub"1
n+1
+class="stub"2
n+1
+class="stub"3
n+1
+…+class="stub"n
n+1
=
n(n+1)
2(n+1)
=class="stub"n
2

数列{bn}={class="stub"1
anan+1
}
的通项公式为bn=class="stub"1
anan+1
=class="stub"2
n
•class="stub"2
n+1
=4(class="stub"1
n
-class="stub"1
n+1

其前n项的和为4[(class="stub"1
1
-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)+…+(class="stub"1
n
-class="stub"1
n+1
)]=4(1-class="stub"1
n+1
)

故选A

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