# Thread: Look at the recursive

1. ## Look at the recursive

I´m having a hard time getting through these sum understanding exercises
This one in particular, If you have some idéa, I value some explaining more then solution so please no: Hey stupid, what the hell, just to the "inversetriangular root extraction recursive absolute value differential" method and you got it!! That just makes me even more confused when I dont understand what I am suppose to do.
The troublesome exercise is:
Look at the series of:

$[a_n]_{n=1}^{\infty}$ of

$a_{n+1}=\frac{5a_n+3}{a_n+3}$ where $1\leq {n}$
with $a_1>0$

and $[b_n]_{n=1}^{\infty}$ Show that $b_n$is a geometric series with $\frac{b_n}{b_{n+1}} =$ Constant when $b_n$ =:
$b_n=\frac{a_n-3}{a_n+1}$ where $1\leq{n}$
Here is my current confused work:
Well I figured lets try some different values on $a_1$ and se what happens, then clearly $a_{n+1}$ goes towards 3. Well that info could be usefull (I suppose a little sometimes). graph looks like $\frac{1}{x}$ graph shiften left 3 units and up 5 units.
Now lets get $a_n$ on one side and everything else on another and se what happens:
so:
1) $a_{n+1}= \frac{5a_n+3}{a_n+3}$

2) $a_{n+1}(a_n+3) = 5a_n+3$

3) $a_{n+1}*a_n+3a_{n+1} = 5a_n +3$

4) $5a_n-a_{n+1}*a_n = 3a_{n+1}-3$

5) $a_n(5-a_{n+1}) = 3a_{n+1}-3$

6) $a_n= \frac{3a_{n+1}-3}{5-a_{n+1}}$
if I set $a_{n+1} = 3$ I get $a_n =3$
but if I set $a_n = 3$ into the $b_n$ formula it goes
$\frac{a_n-3}{a_n+1}= b_n$ = $\frac{3-3}{3+1} = \frac{0}{4}$
And here I am not sure what I´ve done or what I shouldn´t have done...

2. If you solve for $a_n$ you get

$
a_n = - \frac{b_n + 3}{b_n - 1}
$

now substitute this into

$
a_{n+1} = \frac{5a_n+3}{a_n + 3}
$

and simplify and see what happens.

3. Originally Posted by Danny
If you solve for $a_n$ you get

$
a_n = - \frac{b_n + 3}{b_n - 1}
$

now substitute this into

$
a_{n+1} = \frac{5a_n+3}{a_n + 3}
$

and simplify and see what happens.
Yes I get: $a_{n+1} = \frac{b_n+9}{3-b_n}$

4. Originally Posted by Henryt999
Yes I get: $a_{n+1} = \frac{b_n+9}{3-b_n}$
Yes, but also from

$
a_{n} = - \frac{b_{n} + 3}{b_{n} - 1}$

you get

$
a_{n+1} = - \frac{b_{n+1} + 3}{b_{n+1} - 1}$

which you use.

5. Originally Posted by Danny
Yes, but also from

$
a_{n} = - \frac{b_{n} + 3}{b_{n} - 1}$

you get

$
a_{n+1} = - \frac{b_{n+1} + 3}{b_{n+1} - 1}$

which you use.
Im sorry How do you get that?

6. He subbed in the original formula for $a_{n+1}$ and solved for $a_n$

Also, He just incremented the index by 1.
Note: that if $a_n = b_n$ then $a_{n+1} = b_{n+1}$

7. Originally Posted by Haven
He subbed in the original formula for $a_{n+1}$ and solved for $a_n$

Also, He just incremented the index by 1.
Note: that if $a_n = b_n$ for all $n \in \mathbb{N}$ then $a_{n+1} = b_{n+1}$
Small fix!