Find out the answer to "I have to solve this recurrence using substitutions:(n+1)(n−2)an=n(n2−n−1)an−1−(n−1)3an−2..."

Oxinailelpels3t14

Answered question

2022-04-01

I have to solve this recurrence using substitutions:

(n + 1) (n - 2) a_{n} = n (n^{2} - n - 1) a_{n - 1} - {(n - 1)}^{3} a_{n - 2}

with

a_{2} = a_{3} = 1

Answer & Explanation

disolutoxz61

Beginner2022-04-02Added 12 answers

it seems that I solved.
$(n - 2) b_{n} = (n^{2} - n - 1) b_{n - 1} - {(n - 1)}^{2} b_{n - 2}$
So divide it by $n - 1$ then we get
$(1 - \frac{1}{n - 1}) b_{n}$

$= (n - \frac{1}{n - 1}) b_{n - 1} - (n - 1) b_{n - 2}$
$= (n - 1 + 1 - \frac{1}{n - 1}) b_{n - 1} - (n - 1) b_{n - 2}$
$= (n - 1) (b_{n - 1} - b_{n - 2}) + b_{n - 1} (1 - \frac{1}{n - 1})$
then
$(1 - \frac{1}{n - 1}) (b_{n} - b_{n - 1}) = (n - 1) (b_{n - 1} - b_{n - 2})$
then we make subtitution $p_{n} = b_{n} - b_{n - 1}$ as $b_{2} = 3, b_{3} = 4$ we get that $p_{3} = 1$
we have that
$(1 - \frac{1}{n - 1}) p_{n} = (n - 1) p_{n - 1} \Rightarrow p_{n} = \frac{{(n - 1)}^{2}}{n - 2} p_{n - 1} \Rightarrow$
$p_{n} = \prod_{k = 4}^{n} {\frac{{(k - 1)}^{2}}{k - 2}} = \frac{n - 1}{2} \frac{(n - 1)!}{2} \Rightarrow$
$b_{n} - b_{n - 1} = \frac{n! - (n - 1)!}{4} \Rightarrow$
$b_{n} = 4 + \sum_{k = 4}^{n} {\frac{k! - (k - 1)!}{4}} = 4 + \frac{1}{4} (n! - 6) = \frac{10 + n!}{4} \Rightarrow$
$a_{n} = \frac{10 + n!}{4 (n + 1)}$

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