S(n+3)=(n+3)a1+(n+3)(n+2)d1/2
Tn=nb1+n(n-1)d2/2
化简得
na1+3a1-nb1+n^2d1/2+5nd1/2+3d1-n^2d2/2+nd2/2=0
n的系数相同才可约为0：
即
n^2(d1-d2)/2=0
3a1+3d1=0
a1+5d1/2=b1-d2/2
2d1=2d2=-2a1=b1.
不妨取a1=-1,b1=2,d1=d2=1,
an=n-2,bn=n+1即是满足条件的{an}和{bn}.