\(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 $a_3.$ \)

Guest Mar 9, 2019

#1**+1 **

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

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

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

.Rom Mar 9, 2019