等差数列{an}公差为d(d≠0),前n项和
Sn=na1+n(n-1)d/2
∴xn=Sn/n=a1+(n-1)d/2
∴{xn}为等差数列,首项为a1公差为d/2
∴｛Xn｝的前n项和
Tn=n[2a1+(n-1)d/2]/2=na1+n(n-1)d/4
Sn+1-Tn+1=(n+1)nd/2-(n+1)nd/4=(n+1)nd/4
∴1/[S(n+1)-T(n+1)]=4/d*1/[n(n+1)]
=4/d[1/n-1/(n+1)]
∴An=4/d[ (1/1-1/2)+(1/2-1/3)+.+(1/n)+1/(n+1)]
=4/d*[1-1/(n+1)]=4n/[(n+1)d]