And because (xn, yn) is on the curve, yn = xn 2+2.
By the slope formula k=(y 1-y2)/(x 1-x2)
get 2xn =(xn ^ 2+2)/(xn-xn+ 1)
Simplified to xn+1= (xn 2+2)/(2xn)
(2) suppose Xn+ 1-√ 2.
Get xn & gt√2
That is, as long as xn >； is proved; √2, the original inequality holds.
Prove:
xn+ 1=(xn^2+2)/(2xn)=(xn/2)+( 1/xn)
Because x1= 2 >; 0, xn is always greater than 0.
Xn+ 1=(xn/2)+(2/xn)>=2√(xn/2)*( 1/xn)=√2
That is, xn & gt√2 proposition proof
(3)xn+ 1-xn =(xn/2)+( 1/xn)-xn =(2-xn 2)/2xn
Because xn > 2, xn 2 > 2.
That is, xn+ 1-xn < 0, that is, xn monotonically decreases.
Xn is always less than or equal to 2 because x 1=2.
That is, the range of a is (2, positive infinity)