Recursive Formula

**stones44** · March 10th 2008, 04:53 PM

Given: asub0 = 1 asub1 = 3

(asubn)^2 - asubn-1 * asubn+1 = (-1)^n for n greater or equal to 1

find asub3

i simplified it to (square root (n+(asubn-1)*(asubn+1) )= asubn

and i got asub2 = 8 and asub3 = 20.666...

**topsquark** · March 11th 2008, 10:23 AM

Originally Posted by stones44

Given: asub0 = 1 asub1 = 3

(asubn)^2 - asubn-1 * asubn+1 = (-1)^n for n greater or equal to 1

find asub3

i simplified it to (square root (n+(asubn-1)*(asubn+1) )= asubn

and i got asub2 = 8 and asub3 = 20.666...

$a_n^2 - a_{n - 1}a_{n + 1} = (-1)^n$
$a_0 = 1$ and $a_1 = 3$

Am I reading something wrong? I get
$a_n = \frac{a_{n - 1}^2 + (-1)^n}{a_{n - 2}}$

(Hint: Solve the problem for a_{n + 1}, not a_n.)

-Dan

**Soroban** · March 11th 2008, 03:52 PM

Hello, stones44!

It is a rather confusing recurrence.
It took a while to straighten it out . . .

Given: . $a_0 = 1,\;\;\;a_1 = 3.\;\;\;(a_n)^2 - (a_{n-1})\cdot(a_{n+1}) \:= \<img src=$ -1)^n\quad\text{for }n \geq 1" alt="a_0 = 1,\;\;\;a_1 = 3.\;\;\;(a_n)^2 - (a_{n-1})\cdot(a_{n+1}) \:= \

-1)^n\quad\text{for }n \geq 1" />

$\text{Find }\:a_3$

Solve for $a_{n+1}\!:\;\;a_{n+1} \;=\;\frac{(a_n)^2 - (-1)^n}{a_{n-1}}$

For $n = 1$ , we have: . $a_2 \;=\;\frac{(a_1)^2 - (-1)^1}{a_0} \;=\;\frac{3^2 +1}{1}\quad\Rightarrow\quad a_2 \:=\:10$

For $n=2$ , we have: . $a_3 \;=\;\frac{(a_2)^2 - (-1)^2}{a_1} \;=\;\frac{10^2 - 1}{3} \quad\Rightarrow\quad\boxed{a_3 \:=\:33}$

**stones44** · March 12th 2008, 05:35 PM

can you write out the algebra from my step to isolating asub n+1

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