Solution: a=3, or a=-4, …(2 points)
If a=3, then f(x)=x3-3x2+3x- 1, f ′ (x) = 3x2-6x+3 = 3 (x-1) 2 ≥ 0 holds, which satisfies the conditions.
If a=-4, then f(x)=x3-3x2-4x+6,
F'(x)= 3 x2-6x-4 is positive and negative on R,
The condition of "increasing function on R" is not satisfied, so it is abandoned.
So, A = 3...(6 points)
(ii) If f(x)=(x- 1)3, then f(xn)=(xn- 1)3,
Its derivative is f'(x)= 3(x- 1)2,
Let Pn(xn, f(xn))(n∈N+) be the tangent equation of the image with function y=f(x).
For: y-(xn-1) 3 = 3 (xn-1) 2 (x-xn), …(8 points)
Let y = 0:-(xn-1) 3 = 3 (xn-1) 2 (xn+1-xn),
∫xn > 1，
∴xn+ 1=23xn+ 13,xn+ 1？ 1=23(xn？ 1),
∴ Sequence {xn- 1} is a geometric series with 1 as the first term and 23 as the common ratio ... (12 points).
xn- 1=(23)n？ 1, then xn = 1+(23) n? 1 ...( 14 points)