已知等差数列{an}的首项为4,公差为4,其前n项和为Sn,则数列 {1Sn}的前n项和为(  ) A.n2(n+1) B.12n(n+1) C.2n(n+1) D.2nn+1

问题描述:

已知等差数列{an}的首项为4,公差为4,其前n项和为Sn,则数列 {

1
Sn
}的前n项和为(  )
A.
n
2(n+1)

B.
1
2n(n+1)

C.
2
n(n+1)

D.
2n
n+1

∵Sn=4n+

n(n−1)
2
×4=2n2+2n,
1
Sn
1
2n2+2n
1
2
(
1
n
1
n+1
)

∴数列 {
1
Sn
}的前n项和=
1
2
[(1−
1
2
)+(
1
2
1
3
)+…+(
1
n
1
n+1
)]
=
1
2
(1−
1
n+1
)
=
2(n+1)

故选A.