Question

已知数列〔an〕是等差数列,其前n项和为Sn,a3=6,S3=12?(1)求数列〔an〕的通项公式?(2)求1/S1+1

Answer 1

1.

S3=3a3-3d=12,

a3-d=4

d=2

a2=4,a1=2

an=2n

2.

Sn=2n+n(n-1)=n²+n=n(n+1)

1/Sn=1/n-1/(n+1)

1/S1+1/S2+······+1/Sn

=1-1/2+1/2-1/3+1/3-1/4+······+1/n-1/(n+1)

=1-1/(n+1)

=n/(n+1)