I have to solve this recurrence using substitutions:

Oxinailelpels3t14

Oxinailelpels3t14

Answered question

2022-04-01

I have to solve this recurrence using substitutions:
(n+1)(n2)an=n(n2n1)an1(n1)3an2 with a2=a3=1

Answer & Explanation

disolutoxz61

disolutoxz61

Beginner2022-04-02Added 12 answers

it seems that I solved.
(n2)bn=(n2n1)bn1(n1)2bn2
So divide it by n1 then we get
(1-1n-1)bn

=(n-1n-1)bn-1-(n-1)bn-2
=(n1+11n1)bn1(n1)bn2
=(n1)(bn1bn2)+bn1(11n1)
then
(11n1)(bnbn1)=(n1)(bn1bn2)
then we make subtitution pn=bnbn1 as b2=3, b3=4 we get that p3=1
we have that
(11n1)pn=(n1)pn1pn=(n1)2n2pn1
pn=k=4n{(k1)2k2}=n12(n1)!2
bnbn1=n!(n1)!4
bn=4+k=4n{k!(k1)!4}=4+14(n!6)=10+n!4
an=10+n!4(n+1)

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