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Consider the linear recurrence $a_n = 2a_{n−1} − a_{n−2}$ with initial conditions $a_1 = 3, a_0 = 0$. We have

$x^2 − 2x + 1 = (x − 1)^2$.

Thus $ x = 1$ and $a_n$ = $u(1)^n + v(1)^n$.

Why do we get $a_n = u + nv$ instead of $a_n = u + v$ since $1^n = 1$?

JohnD
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Gavin
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    When you have repeated roots (multiplicity more than $1$), the form of the generic solution changes. Check http://math.stackexchange.com/questions/129849/linear-homogeneous-recurrence-relations-with-repeated-roots-motivation-behind-l – Macavity Oct 28 '14 at 10:16
  • it must be $a_n=A+nB$ – Dr. Sonnhard Graubner Oct 28 '14 at 10:48

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$x=1$ is a repeated root of your characteristic equation, so the general solution is $$a_n=u\cdot(1)^n+v\cdot n(1)^n=u+nv.$$

Applying the initial conditions, $$ a_0=u+0 \cdot v=0\implies u=0 \quad\text{ and }\quad a_1=u+1\cdot v=3\implies v=3 $$ so the solution is $a_n=3n$, $n=0,1,2,\dots$

JohnD
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