a(n+1)=pa(n)+q

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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