分类：教育/科学-理工学科-数学
问题：已知数列{an}是等差数列，其前n项和为Sn，a3=6，S3=1? 
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
请写过程 谢谢
回答：数列{an}是等差数列，其前n项和为Sn，a3=6，S3=12;
1)
a3 = s3 - s2, s2 = s3-a3 = 6;
a1 + a2 = 6;
a3 = 6;
公差d = a3 -a2 = a1;
a3 = a1 + 2d = 3a1;
a1 = 2, d = 2;
通项为：
an = a1 + (n-1)d
   = 2 + 2(n-1)
   = 2n;

2)
sn = a1+a2+...+ an
   = 2+2*2 +...+2n
   = 2*(1+2+...+n)
   = 2*(1+n)*n/2
   = (1+n)*n

1/sn = 1/(n*(1+n)) = 1/n-1/(n+1)
1/s1+1/s2+...+1/sn 
=(1-1/2)+(1/2-1/3) +...+ (1/n-1/(n+1))
=1-1/(n+1)
=n/(n+1).