Question

已知{an}是等差数列其前n项和为sn,a3=6,s3=12求{an}的通项公式求证1/s1+1/s2+...1/sn<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)

Answer 2

一、S3=a1+a2+a3=3a3-3d=12,
d=2、a2=4,a1=2 an=a1+（n-1）d=2n
二、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)
=n/(n+1)

Answer 3

1） s3= 12 = 3 a2
a2 =4 ；d=a3-a2 =2;a1 =2
an =2n
2) Sn=n(a1 + an) /2=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)