设等差数列an的前n项和为sn,已知a3=4,s3=9.(1)求数列an的通项公式;(2)令bn=an乘an+1分之1,

a3=a1+2d=4;
s3=a3+a2+a1=3a1+3d=9; 推得a1=2,d=1;
所以{an}的通项：an=2+(n-1)=n+1;
bn=1/(an.a(n+1))=1/((n+1)(n+2))=1/(n+1)-1/(n+2);
b1=1/6;
{bn}的前n项和：sbn=(b1+bn).n/2=n/2.(1/6+1/(n+1)-1/(n+2));