1. ## sequence problem

given the sequence :

$x_{n+1} =\frac{6x_{n}+1}{2x_{n}+5}$

find its general term

2. Originally Posted by xalk
given the sequence :

$x_{n+1} =\frac{6x_{n}+1}{2x_{n}+5}$

find its general term
Hi

If $x_0 = -\frac12$ then $\forall n~
x_n = -\frac12$

If $x_0 \neq -\frac12$ then let $y_n = \frac{x_n - 1}{2\:x_n + 1}$

You can show that $y_{n+1} = \frac47\:y_n$ leading to $y_n = \alpha\:\left(\frac47\right)^n$ where $\alpha = y_0 = \frac{x_0 - 1}{2\:x_0 + 1}$

Spoiler:

And finally $x_n = \frac{7^n + \alpha\:4^n}{7^n - 2 \alpha\:4^n}$

When $\alpha \rightarrow \pm\infty$ $x_n \rightarrow -\frac12$

3. Originally Posted by running-gag
Hi

If $x_0 = -\frac12$ then $\forall n~
x_n = -\frac12$

If $x_0 \neq -\frac12$ then let $y_n = \frac{x_n - 1}{2\:x_n + 1}$

You can show that $y_{n+1} = \frac47\:y_n$ leading to $y_n = \alpha\:\left(\frac47\right)^n$ where $\alpha = y_0 = \frac{x_0 - 1}{2\:x_0 + 1}$

Spoiler:

And finally $x_n = \frac{7^n + \alpha\:4^n}{7^n - 2 \alpha\:4^n}$

When $\alpha \rightarrow \pm\infty$ $x_n \rightarrow -\frac12$
THANK you,suppose the 1st member is 3 what is the limit of the sequence.

I am a Little confused

4. Do you mean $x_0 = 3$ ?

Then $\alpha = \frac{3 - 1}{2\cdot3 + 1} = \frac27$

$x_n = \frac{7^n + \alpha\:4^n}{7^n - 2 \alpha\:4^n} = \frac{1 + \alpha\:\left(\frac47\right)^n}{1 - 2 \alpha\:\left(\frac47\right)^n}$

Therefore $\lim_{n\rightarrow +\infty} x_n = 1$