Oxinailelpels3t14

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 ${a}_{2}={a}_{3}=1$

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 ${b}_{2}=3,\text{}{b}_{3}=4$ we get that ${p}_{3}=1$

we have that

$(1-\frac{1}{n-1}){p}_{n}=(n-1){p}_{n-1}\Rightarrow {p}_{n}=\frac{{(n-1)}^{2}}{n-2}{p}_{n-1}\Rightarrow$

${p}_{n}=\prod _{k=4}^{n}\left\{\frac{{(k-1)}^{2}}{k-2}\right\}=\frac{n-1}{2}\frac{(n-1)!}{2}\Rightarrow$

${b}_{n}-{b}_{n-1}=\frac{n!-(n-1)!}{4}\Rightarrow$

${b}_{n}=4+\sum _{k=4}^{n}\left\{\frac{k!-(k-1)!}{4}\right\}=4+\frac{1}{4}(n!-6)=\frac{10+n!}{4}\Rightarrow$

$a}_{n}=\frac{10+n!}{4(n+1)$

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