已知数列{an}中,a1=a>0,an+1=f(an),其中f(x)=3x/1—+2x.(1)求a2,a3,a4.(2)猜想｛an｝通项公式

a(n+1)=3a(n)/[1+2a(n)],若a(n+1)=0,则,a(n)=0, ..., a(1)=0,与a(1)=a>0矛盾.因此,a(n)不为0.1/a(n+1)=[1+2a(n)]/[3a(n)] = (1/3)[1/a(n)] + 2/3 = (1/3)[1/a(n)] + 1 - 1/3,1/a(n+1) - 1 = (1/3)[1/a(n) - 1],{1/a(...