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\(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
avatar+6042 
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\(\text{interpreting this mangled bit of text I get}\\ a_n^2 - a_{n-1} a_{n+1}=(-1)^n,~n\in \mathbb{N}\\ a_0=1,~a_1=5\)

 

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

 

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

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 Mar 9, 2019
 #2
avatar+7712 
+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

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