若(1+x)n+1的展开式中含xn-1的系数为an,则1a1+1a2+…+1an的值为()A.nn+1B.2nn+1C.n(n+1)2D.n(n+3)2-数学

题目简介

若(1+x)n+1的展开式中含xn-1的系数为an,则1a1+1a2+…+1an的值为()A.nn+1B.2nn+1C.n(n+1)2D.n(n+3)2-数学

题目详情

若(1+x)n+1的展开式中含xn-1的系数为an,则
1
a1
+
1
a2
+…+
1
an
的值为(  )
A.
n
n+1
B.
2n
n+1
C.
n(n+1)
2
D.
n(n+3)
2
题型:单选题难度:偏易来源:不详

答案

由题意可得
an=Cn+1n-112=Cn+12=
(n+1)•n
2
,∴class="stub"1
an
=class="stub"2
n(n+1)
=2•(class="stub"1
n
-class="stub"1
n+1
),
class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
an

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

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

故选项为B

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