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。S3=3a3-3d=12,a3-d=4d=2a2=4,a1=2an=2n2。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)