对于正项数列{an},定义Hn=na1+2a2+3a3+…+nan为{an}的“给力”值,现知某数列的“给力”值为Hn=2n+2,则数列{an}的通项公式为an=(  ) A.12n+1 B.1n+1 C.12+n D.2n-12

问题描述:

对于正项数列{an},定义Hn=

n
a1+2a2+3a3+…+nan
为{an}的“给力”值,现知某数列的“给力”值为Hn=
2
n+2
,则数列{an}的通项公式为an=(  )
A.
1
2n
+1

B.
1
n
+1

C.
1
2
+n

D. 2n-
1
2

根据题意,得;

n
a1+2a2+3a3+…+nan
=
2
n+2

∴a1+2a2+3a3+…+nan=
n(n+2)
2

∴a1+2a2+3a3+…+(n-1)an-1=
(n-1)(n+1)
2

两式相减,得nan=
n(n+2)-(n-1)(n+1)
2

∴an=
2n+1
2n
=1+
1
2n

故选:A.