Having this recurrence relation
$a_n = 5a_{n-1} - 6a_{n-2} + 4·3^n$
$a_1 = 36$
$a_0 = 0$
How can I solve this?
I tried by characteristics roots and got stuck:
*making $a_n=r^n$
$r^n = 5r^{n-1} - 6r^{n-2} - 4·3^n$
*dividing by $r^{n-2}$ and equating to zero got:
$r^2 + 6 - 5r - 4·3^2 = 0$
*solving the equation and getting the roots got an irrational root:
$= {-5+-\sqrt{145}\over2} $
I'm stuck there,
How can I solve this with the irrational roots?
What other methods can I try?
Thanks
The sequence in question is \begin{align} a_{n+2} - 5 a_{n+1} + 6 a_{n} = 4 \cdot 3^{n+2} \end{align} where $a_{0} = 0$ and $a_{1} = 36$. Considering the primary equation $a_{n+2} - 5 a_{n+1} + 6 a_{n} =0$ for which, by taking $a_{n} \approx r^{n}$, leads to $(r-2)(r-3)= 0$ and hence $a_{n} = A 2^{n} + B 3^{n}$. Now for the particular solution it is noticed that the term $3^{n}$ appears in the homogenous solution and particular. Due to this then take $C \, 3^{n} n$ as the particular solution. From this the general form is now \begin{align} a_{n} = A 2^{n} + B 3^{n} + C 3^{n} n. \end{align}
Applying the initial condition $a_{0} = 0$ it is seen that $B = -A$ and \begin{align} a_{n} = A(2^{n} - 3^{n}) + C 3^{n} n. \end{align} Applying this solution to the equation in question it is determined that $C = 12$ and leads to \begin{align} a_{n} = A(2^{n} - 3^{n} ) + 4 \cdot 3^{n+1} \cdot n. \end{align} Applying the condition $a_{1} = 36$ leads to $A = 0$ and hence $a_{n} = 4 \cdot 3^{n+1} \cdot n$.