+0

# I don't really understand this at all

+1
124
2

$$Given a_0 = 1 and a_1 = 5, and the general relation $a_n^2 - a_{n - 1} a_{n + 1} = (-1)^n$for n \ge 1, find a_3.$$

Mar 9, 2019

#1
+5655
+1

$$\text{interpreting this mangled bit of text I get}\\ a_n^2 - a_{n-1} a_{n+1}=(-1)^n,~n\in \mathbb{N}\\ a_0=1,~a_1=5$$

$$\text{one would think we'd like to calculate } a_2 \text{ next}\\ \text{applying the rule with }n=1 \text{ we have}\\ a_1^2 - a_0 a_2 = (-1)^1 = -1\\ 5^2 - (1) a_2 = -1\\ a_2 = 26$$

$$\text{Now just repeat that with }n=2 \text{ to solve for }a_3$$

.
Mar 9, 2019
#2
+7709
+1

It is sequence A052918 on OEIS. a3 is 5(26) + 5 = 135 as the transition formula can be rewritten as $$a_{n+1} = 5a_n + a_{n-1}$$.

https://oeis.org/A052918 <-- Here's a link to the page about sequence on OEIS.

Mar 10, 2019