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 {
}的前n项和为( )1 Sn
A.
n 2(n+1)
B.
1 2n(n+1)
C.
2 n(n+1)
D.
2n n+1
答
∵Sn=4n+
×4=2n2+2n,n(n−1) 2
∴
=1 Sn
=1 2n2+2n
(1 2
−1 n
).1 n+1
∴数列 {
}的前n项和=1 Sn
[(1−1 2
)+(1 2
−1 2
)+…+(1 3
−1 n
)]=1 n+1
(1−1 2
)=1 n+1
.n 2(n+1)
故选A.