(1) Personally, I think the condition of a 1=3/5 is useless. I tried a 1=6/5, and the result was the same. Bn is a arithmetic progression with an error of 1, because this condition is not used at all in my proof.

Find f(x) first. The original function is shifted to the right by 2, 1 to get f(x), so f(x) is shifted to the left by 2, 1 to get the original function, that is, (x-2, f(x)- 1) is on the original function. F (x)-1=1-1((x-2)+2), that is, f (x) = 2-1/x.

Therefore, an=2- 1/an- 1, that is, an-1/an-1,that is,1/(an-1) =

Therefore, bring BN-BN-1= BN-1* (an-1-1) into BN-BN-1=1(an-/kloc-. So this is not a condition. .

(2) if bn-bn- 1 = 1 and b 1=-5/2, it is easy to know that bn = n-7/2, then an =1+1(n-7/2

Make a < 0 get 5/2.

The rest of the time, when n & gt=4, an is always positive, and it obviously decreases with the increase of n, so it is easy to know an >: 1, while a 1=3/5, a2= 1/3, a4=3, an.

Therefore, the minimum value a3=- 1 and the maximum value a4=3.