f(x)=(2x+3)/(3x),则an+1=f(1/an),得a(n+1)=a(n)+2/3,又a1=1,所以a(n)=1+(n-1)2/3;a(2n)a(2n-1)=[1+(2n-1)2/3][1+(2n-2)2/3]=16n^2/9-1/9;T(n)=-4/3[a(2)+a(4)+……+a(2n)]=-4/3(5/3+4n/3+1/3)n/2=2/3(4n/3+2)n...