a(n+1)=pa(n)+q can be transformed into a(n+1)-c=p(a(n)-c).
a(n+1)-c=p(a(n)-c)
⇔
a(n+1)=pa(n)-pc+c
comparing with a(n+1)=pa(n)+q,
we get q=-pc+c ⇔ c=pc+q.
as a result, in a(n+1)=pa(n)+q,
we can think of a(n+1) and a(n) as c,
then we get c=pc+q.
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