written 7.9 years ago by | modified 2.8 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Discrete Structures
Marks: 8 Marks
Year: May 2015
written 7.9 years ago by | modified 2.8 years ago by |
Mumbai University > Computer Engineering > Sem 3 > Discrete Structures
Marks: 8 Marks
Year: May 2015
written 7.9 years ago by |
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
$a_n =20 – 47n + 22n^2$