S3 = 3a 1+3d = a4+6 3a 1+3d = a 1+3d+6 a 1 = 3，d=2

an = a 1+(n- 1)d = 2n+ 1

(2)sn=na 1+n(n- 1)d/2=3n+n(n- 1)=n^2+2n

1/sn= 1/(n^2+2n)= 1/n(n+2)= 1/2*[(n+2)-n]/[n*(n+2)]= 1/2[ 1/n- 1/(n+2)]

Let Bn = 1/sn and cn be the sum of the first n terms of Bn;

2Cn = 2 *( 1/s 1+ 1/S2+ 1/S3+……+ 1/Sn)=( 1/ 1- 1/3)+( 1/3- 1/5)+( 1/5- 1/7)+

So the demand is Cn=(n+ 1)/2(n+2).

This is a relatively basic topic. You should read the formula by yourself, see how to use the formula in the textbook, and check the exercises after class. You'd better do it yourself.