数学补充:等差数列{an}中,a1=3,d=2,Sn为前n项和,求1/S1+1/S2+.+1/Sn的值.求详解,要步骤.

a1=3,d=2
∴ an=a1+(n-1)d=3+2(n-1)=2n+1
∴ Sn=(a1+an)*n/2=(3+2n+1)*n/2=n²+2n
∴ 1/Sn=1/[n(n+2)]=(1/2)[1/n-1/(n+2)]
∴ 1/S1+1/S2+.+1/Sn
=(1/2)[1-1/3+1/2-1/4+1/3-1/5+.+1/(n-1)-1/(n+1)+1/n-1/(n+2)]
=(1/2)[1+1/2-1/(n+1)-1/(n+2)]
=(3/4)-(1/2)/(n+1)-(1/2)/(n+2)