等差数列{an}.a1=3,d=2,Sn为前n项和,求S=1/s1+1/s2+.+1/sn
问题描述：
等差数列{an}.a1=3,d=2,Sn为前n项和,求S=1/s1+1/s2+.+1/sn
答
an=3+2（n-1）=2n+13+5+……+(2n+1)=(3+2n+1)*n/2=n(n+2)所以1/[3+5+……+(2n+1)]=1/n(n+2)=(1/2)*[1/n-1/(n+2)]S=1/s1+1/s2+.+1/sn=1/2*[1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+……+1/(n-1) - 1/(n+1)+1/n - 1/(n+2)]=1/2...