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Find the solution to the recurrence relation $a_n=6a_{n-1}+11a_{n-2}-6a_{n-3}$ given $a_0=20,a_1=5 \ \ and \ \ a_2=15$

Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 8 Marks

Year: May 2015

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The characteristic equation of the recurrence relation is $r^3-6r^2-11r+6=0$

Its roots are $r=-1$. Hence the sequence {$a_n$} is a solution to the recurrence relation if and only if

$a_n=α_1 (-1)^n+-α_2 (n)(-1)^n+α_3 (n \times n)(-1)^n=α_1+α_1n+α_3(n \times n)$

For some constant $α_1$ and $α_2$.

From the initial condition, it follows that

$a_0=20=α_1 + α_2(0) + α_3(0) \\ a_1= 5 = - α_1 - α_2 - α_3 \\ a_2= 15 = α_1 +2α_2 + 4 α_3$

Solving the equations, we get $α_1=20, α_2=-47, α_3=22$

Hence the solution is the sequence {$a_n$} with

$a_n =20 – 47n + 22n^2$

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