Originally Posted by
HallsofIvy You mean "xn", not "xD". The first thing I would do is calculate a few terms. [tex]x_1= 5[/tex[ so $\displaystyle x_2= 2x_1- 5= 2(5)- 5= 10- 5= 5$. Then $\displaystyle x_3= 2x_2- 5= 2(5)- 5= 5$. Then $\displaystyle x_4= 2x_3- 5= 2(5)- 5= 10- 5= 5$. See a pattern? Let's make the "guess" that, for all n, $\displaystyle x_n= 5$ and prove it by "induction".
We are given that $\displaystyle x_1= 5$. Suppose that, for some k, $\displaystyle x_k= 5$. Then [tex]x_{k+1}= 2x_k- 5= 2(5)- 5= 10- 5= 5! By "induction", [tex]x_n= 5[/rex] for all n. "Induction", here, allows us to extend the fact that, at each step we are doing the same thing, to all positive integers.