Then PQn passes (Xn+ 1
, 0) point.
That is, (f(Xn)-5)/(Xn-4)(Xn+ 1
-4)+5=0, brought into the calculation are
Xn+ 1=4-5/(xn
+2),
+1 both sides,
Xn+ 1
+ 1
=
5(xn
+ 1)/(xn
+2), in turn, make yn= 1/(xn
+ 1);
Get 5*
yn+ 1
= 1
+
yn;
Subtract 5/4 from both sides, there is
5(yn+ 1
- 1/4)=yn
- 1/4;
Yn= 1/(xn+ 1),y 1- 1/4 = 1/4 = 1/ 12;
yn = 1/ 12 *( 1/5)(n- 1)+ 1/4;
The general formula of xn is
xn=[9-( 1/5)^(n- 1)]/[( 1/5)^(n- 1)+3]; (n=2 can be brought in for checking, and x2 = (44/5)/(16/5) =1/4 satisfies)
Only the second question is written here, but this common term proves that the first question is also indispensable.
The first question is thinking.
4-5/(x+2)>x, at x∈(- 1, 3), is bound to hold, which proves that both sides are sufficient, x < 3, x>2. Get a license.