Question

25. Solve recurrence relation an+3=3an+2+4an+1-12an for n20 with a0-0,al--11,a2--15

Answer 1

Let us solve recurrence relation $a_{n+3}=3a_{n+2}+4a_{n+1}-12a_n$ with $a_0=0,a_1=11,a_2=15.$ Firstly, let us solve its characteristic equation $k^3=3k^2+4k-12,$ which is equivalent to $(k-2)(k^2-k-6)=0,$ and hence to $(k-2)(k+2)(k-3)=0.$ We conclude that its roots are $k_1=2,\ k_2=-2,\ k_3=3.$ The general solution is $a_n=c_12^n+c_2(-2)^n+c_33^n.$ It follows that $0=a_0=c_1+c_2+c_3,\ 11=a_1=2c_1-2c_2+3c_3,\ 15=a_2=4c_1+4c_2+9c_3.$ The system $\begin{cases}c_1+c_2+c_3=0\\ 2c_1-2c_2+3c_3=11\\ 4c_1+4c_2+9c_3=15\end{cases}$ is equivalent to the system $\begin{cases}c_1+c_2+c_3=0\\ -4c_2+c_3=11\\ 5c_3=15\end{cases},$ and hence

$\begin{cases}c_1=-1\\ c_2=-2\\ c_3=3\end{cases}.$

Therefore, the solution is $a_n=-2^n-2(-2)^n+3\cdot3^n=-2^n+(-2)^{n+1}+3^{n+1}.$

If $n=20,$ then $a_{20}=-2^{20}+(-2)^{21}+3^{21}=-2^{20}-2\cdot 2^{20}+3^{21}=3^{21}-3\cdot 2^{20}.$