Is there any way I can make an explicit formula for the sequance $a_n=x+ya_{n-1}$?

Question

Let $a_n$ be a sequence defined by recursion: $a_n=x+ya_{n-1}, a_1=k$. For example, if $(x,y)=(3,5)$, then the sequence would go
$$a=\{k,\space 3+5k,\space 3+5(3+5k),\space ...\}$$ Is there an explicit formula for $a_n$? If not, is there a way to tell if a number is a member of $a$?

score 2 · Answer 1 · answered Apr 26 '20 at 20:50

2

Hint: Let $b_n = a_{n+1}-a_n$. Then $b_n = y b_{n-1}$. Can you finish?

answered Apr 26 '20 at 20:50

lhf

216,483

See https://en.wikipedia.org/wiki/Recurrence_relation#Solving_non-homogeneous_linear_recurrence_relations_with_constant_coefficients – lhf Apr 26 '20 at 20:50
If you can finish, please add a complete answer yourself for future visitors – lhf Apr 26 '20 at 21:21

score 2 · Accepted Answer · answered Apr 27 '20 at 03:24

2

Just another solution.

Considering $$a_n=x+y\,a_{n-1}$$ let $a_n=b_n+k$ and replace $$b_n+k=x +y\ b_{n-1}+k y$$ Let $k=x+k y \implies k=\frac x{1-y}$ (if $y \neq 1$) to make $$b_n=y \,b_{n-1}$$ which is simple.

Solve for $b_n$ and $a_n=b_n+\frac x{1-y}$

answered Apr 27 '20 at 03:24

Claude Leibovici

260,315

(+1) I was just about to add an answer very similar. – robjohn Apr 27 '20 at 03:38
@robjohn. In French, we have an expression "great minds meet". The problem is that I am not ! Cheers and thanks. – Claude Leibovici Apr 27 '20 at 03:44
Silly question: why does $k=x+ky$? – CatPerson Apr 27 '20 at 14:06
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@CatPerson. Just to make the constant term equal to zero – Claude Leibovici Apr 27 '20 at 14:43
@ClaudeLeibovici Why can we let k equal what we want? – CatPerson Apr 27 '20 at 16:35
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@CatPerson. We solve for $k$ such that ........ – Claude Leibovici Apr 27 '20 at 18:50

Is there any way I can make an explicit formula for the sequance $a_n=x+ya_{n-1}$?

2 Answers2