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/S2+·...

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/S2+······+1/Sn的值

Answer 1

a3=a1+2d、S3=3a1+3d 所以将a3=6、S3=12代入得a1=2,d=2故an=2n
Sn=n(n+1),1/S1+1/S2+······+1/Sn=1/(1*2)+1/(2*3)+...+1/n(n+1)\
=1-1/2+1/2-1/3+1/3-...+1/(n-1)-1/n+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)

Answer 2

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)