等差数列an中,a2+a3=7,a4+a5+a6=18,第一问求an的通项公式 ;第二问 设数列an的前n项和为Sn,求

1)设首项为a1,公差为da2+a3=7=2a1+3d,a4+a5+a6=18=3a1+12d得 a1=2 ,d=1an=1+n (n为大于0的整数）2) Sn=na1+n(n-1)d/2=2n+n^2/2-n/2=n(n+3)/23)1/Sn=2/[n(n+3)]=2/3[1/n-1/(n+3)]所以S3+S6+。+S3n=2/3[1/3-1/6+1/6-1/9+。+1/3(n-1)-1/3n]= 2[1/3-1/3n]/3=2(n-1)/9n