已知数列{An}、{Bn}满足a1=1/2 b1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm

(1)A2=A1*A1=1/4,A2/A1=1/2;
An+1=An*A1,An+1/An=A1=1/2;所以An为等比数列,An=1/(2^n).
B2=B1+B1=-1,B2-B1=-1/2;
Bn+1=Bn+B1,Bn+1-Bn=B1=-1/2;所以Bn为等差数列,
Bn=-1/2+(n-1)(-1/2)=-n/2.
(2)记Dn=AnBn,Tn=D1+D2+D3+...+Dn
=-(1/2)*(1/2)-(2/2)*(1/4)-(3/2)*(1/8)-(4/2)*(1/16)-...-(n/2)*(1/2^n);
2Tn==-(1/2)*1-(2/2)*(1/2)-(3/2)*(1/4)-(4/2)*(1/8)-...-(n/2)*(1/2^n-1);
2Tn-Tn=-[1/2+1/4+1/8+1/16+...+1/2^n]+(n/2)*(1/2^n)
=-[1-1/2^n]+(n/2)*(1/2^n)=-1+(n/2+1)*(1/2^n);
即：Tn=-1+(n/2+1)*(1/2^n);(此方法称为错位相减法)
(3)-n/2=(4Cn+n)/(3Cn+n),-3nCn-n^2=8Cn+2n,Cn=-(n^2+2n)/(3n+8);
令3n+8=t,因为n>=1,t>=11;所以：
Cn=-[(1/9)(t^2-16t+64)+2(t-8)/3]/t
=-(t+16/t-10)/9;由于t+16/t当t>=11时是递增的,所以Cn就是递减的,所以t=11时Cn最大,即n=1时Cn最大,所以：Cn