(1)因为该数列为常数列,所以a1=a2=f(a1)由a1=a代人得a=-1或1(2)b(n+1)/bn=。。。(由关系式代入化简可得b(n+1)/bn=1/3)所以bn=b1×(1/3)^(n-1)=(an—1)/(an+1) 化简得an=(3^n+1)/(3^n—1)