1. Finding x1000

Hi there, I am completely stuck and would like some help with this. I’d appreciate it very much:

Given that x1=5 and xD=2xn-1 -5 for n>1. Find x1000

Thanks again!

2. Re: Finding x1000

Originally Posted by Asiram
Hi there, I am completely stuck and would like some help with this. I’d appreciate it very much:

Given that x1=5 and xD=2xn-1 -5 for n>1. Find x1000

$(\forall n)[x_n=5]$ see here.

3. Re: Finding x1000

I think I understand where that came from. How would I use this to find x1000?

Thanks!

4. Re: Finding x1000

Originally Posted by Plato
$(\forall n)[x_n=5]$ see here.
Did you not understand what this meant? $\displaystyle x_{1000}= 5$!

5. Re: Finding x1000

Originally Posted by Asiram
Hi there, I am completely stuck and would like some help with this. I’d appreciate it very much:

Given that x1=5 and xD=2xn-1 -5 for n>1. Find x10

Thanks again!
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.

6. Re: Finding x1000

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.
Yeah, that was a typo. Thank you both for your help. Truly appreciate it!

7. Re: Finding x1000

Originally Posted by Asiram
Given that x1=5 and xD=2xn-1 -5 for n>1. Find x1000

How would it be worded if "properly worded"?
x1 = 5 means 1st term of an arithmetic sequence?

8. Re: Finding x1000

Originally Posted by DenisB
How would it be worded if "properly worded"?
x1 = 5 means 1st term of an arithmetic sequence?
First of all it is clearly not an arithmetic sequence.
It is a recursively defined sequence: the first term is $x_1=5$, then for $n>1,~x_n=2\cdot x_{n-1}-5$.
That is each term after first term, each tern is twice the preceding term minus five.

9. Re: Finding x1000

Thanks for the Tylenol Plato!
I'll send you a 12ouncer of Canadian Club for Xmas...