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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 $a$?

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  • $\begingroup$ Here is a related question. $\endgroup$
    – rtybase
    Apr 27 '20 at 0:10
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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 $a_n=b_n+\frac x{1-y}$

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  • $\begingroup$ (+1) I was just about to add an answer very similar. $\endgroup$
    – robjohn
    Apr 27 '20 at 3:38
  • $\begingroup$ @robjohn. In French, we have an expression "great minds meet". The problem is that I am not ! Cheers and thanks. $\endgroup$ Apr 27 '20 at 3:44
  • $\begingroup$ Silly question: why does $k=x+ky$? $\endgroup$
    – CatPerson
    Apr 27 '20 at 14:06
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    $\begingroup$ @CatPerson. Just to make the constant term equal to zero $\endgroup$ Apr 27 '20 at 14:43
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    $\begingroup$ @CatPerson. We solve for $k$ such that ........ $\endgroup$ Apr 27 '20 at 18:50
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Hint: Let $b_n = a_{n+1}-a_n$. Then $b_n = y b_{n-1}$. Can you finish?

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  • $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$
    – lhf
    Apr 26 '20 at 20:50
  • $\begingroup$ If you can finish, please add a complete answer yourself for future visitors $\endgroup$
    – lhf
    Apr 26 '20 at 21:21

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