在等差数列{an}中,a2+a3=7,a4+a5+a6=18,求通项公式.设前n项和为sn,求1/s3+1/s6+...

a2+a3=7,a4+a5+a6=18
即有：(a1+d)+(a1+2d)=7,即2a1+3d=7
(a1+3d)+(a1+4d)+(a1+5d)=18,即3a1+12d=18,a1+4d=6
解得：d=1,a1=2
故an=a1+(n-1)d=2+n-1=n+1
Sn=(a1+an)n/2=(2+n+1)n/2=n(n+3)/2
1/Sn=2/n(n+3)=2/3*[1/n-1/(n+3)]
1/S3+1/S6+...+1/S3n=2/3[1/3-1/6+1/6-1/9+...+1/3n-1/(3n+3)]=2/3[1/3-1/(3n+3)]=2/3*n/(3n+3)