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A Linear Recurrence Relation is one of the form

u0 is a constant, and
un+1 = a un + b

On this page we'll find a closed form for this type of recurrence relation.

Linear Recurrence Relation

A linear recurrence relation of the form un+1=aun+b has the closed form

un = (u0+b/(a-1))an - b/(a-1)

Here's how to derive it:


u2=au1+b = a(au0+b)+b = u0a2 + ba + b

u3=au2+b = a(u0a2+ba+b)+b = u0a3+ba2+ba+b

u4=au3+b = a(u0a3+ba2+ba+b)+b = u0a4+ba3+ba2+ba+b


un = u0an+ban-1+...+ba2+ba+b

To simplify the right-hand-side, note that

an-1+an-2+...+a+1 = (an-1)/(a-1)


un = u0an + b(an-1)/(a-1)

un = u0an + ban/(a-1) - b/(a-1)

un = (u0+b/(a-1))an-b/(a-1)

Related pages in this website

Fibonacci Numbers -- whose successive ratios approach big phi

Recurrence Relation Puzzle 1, having to do with Fibonacci Numbers

Increasing Function Puzzle 1 -- Find the closed form of this function defined as a recurrence relation


The webmaster and author of this Math Help site is Graeme McRae.