Oxinailelpels3t14

## Answered question

2022-04-01

I have to solve this recurrence using substitutions:
$\left(n+1\right)\left(n-2\right){a}_{n}=n\left({n}^{2}-n-1\right){a}_{n-1}-{\left(n-1\right)}^{3}{a}_{n-2}$ with ${a}_{2}={a}_{3}=1$

### Answer & Explanation

disolutoxz61

Beginner2022-04-02Added 12 answers

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

$=\left(n-\frac{1}{n-1}\right){b}_{n-1}-\left(n-1\right){b}_{n-2}$
$=\left(n-1+1-\frac{1}{n-1}\right){b}_{n-1}-\left(n-1\right){b}_{n-2}$
$=\left(n-1\right)\left({b}_{n-1}-{b}_{n-2}\right)+{b}_{n-1}\left(1-\frac{1}{n-1}\right)$
then
$\left(1-\frac{1}{n-1}\right)\left({b}_{n}-{b}_{n-1}\right)=\left(n-1\right)\left({b}_{n-1}-{b}_{n-2}\right)$
then we make subtitution ${p}_{n}={b}_{n}-{b}_{n-1}$ as  we get that ${p}_{3}=1$
we have that
$\left(1-\frac{1}{n-1}\right){p}_{n}=\left(n-1\right){p}_{n-1}⇒{p}_{n}=\frac{{\left(n-1\right)}^{2}}{n-2}{p}_{n-1}⇒$
${p}_{n}=\prod _{k=4}^{n}\left\{\frac{{\left(k-1\right)}^{2}}{k-2}\right\}=\frac{n-1}{2}\frac{\left(n-1\right)!}{2}⇒$
${b}_{n}-{b}_{n-1}=\frac{n!-\left(n-1\right)!}{4}⇒$
${b}_{n}=4+\sum _{k=4}^{n}\left\{\frac{k!-\left(k-1\right)!}{4}\right\}=4+\frac{1}{4}\left(n!-6\right)=\frac{10+n!}{4}⇒$
${a}_{n}=\frac{10+n!}{4\left(n+1\right)}$

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