设an=m(2n+1),bn=m(3n+1)
Sn=[3m+m(2n+1)]n/2=m(n+2)n
Tn=[4m+m(3n+1)]n/2=m(3n+5)n/2
s11/t11=m（11+2）*11/[m(3*11+5)*11/2]=13/19
s2n-1/t2n-1=m(2n-1+2)(2n-1)/[m(6n-3+5)(2n-1)/2]=(2n+1)/(3n+1)
Sn/Tn=[(a1+an)n/2]/[(b1+bn)n/2]=3n+1/2n+3
可设Sn=m(3n+1)n/2,Tn=m(2n+3)n/2
a1=S1=2m,S2=7m,a2=S2-a1=5m,d1=3m
b1=T1=5m/2,T2=7m,b2=T2-b1=7m-5m/2=9m/2,d2=2m
a5/b5=(2m+3m*4)/(5m/2+2m*4)=4/3
an/bn=[2m+3m*（n-1）]/[5m/2+2m*(n-1)]=(6n-2)/(4n+1)
通过巧设系数m来使运算准确,虽然复杂点,而且易于理解