已知函数f(x)=(2x+1)/(x+2)(x不等于2,x∈R),数列{an}满足a1=t(t不等于-2,t∈R), a(n+1)=f(an)(n∈N)

（1）a(n+1)=f(an)(n∈N)由此可知a(n+1)=(2an+1)/(an+2)当数列为常数列时,有an=a(n+1)带入a(n+1)=(2an+1)/(an+2)解得an=±1,即t=1或-1为所求（2）bn=(an+1)/(an-1) ①b(n+1)=[a(n+1)+1]/[a(n+1)-1] ②将a(n+1)=(2an+...