Question

an=-an-1+16an-2+4an-3-48an-4,where a°=0,a1=16,a2=-2,a3=142

Answer 1

Let us solve $a_n=-a_{n-1}+16a_{n-2}+4a_{n-3}-48a_{n-4},$ where $a_0=0,a_1=16,a_2=-2,a_3=142$.

Firstly, let us solve the characteristic equation

$k^4=-k^3+16k^2+4k-48$

$k^4+k^3-16k^2-4k+48=0$

$k^4-4k^2+k^3-4k-12k^2+48=0$

$k^2(k^2-4)+k(k^2-4)-12(k^2-4)=0$

$(k^2+k-12)(k^2-4)=0$

$(k-3)(k+4)(k-2)(k+2)=0$

It follows that the characteristoc equation has the following roots:

$k_1=3,\ k_2=-4,\ k_3=2,\ k_4=-2.$

Therefore, the general solution is $a_n=c_13^n+c_2(-4)^n+c_32^n+c_4(-2)^n.$

Since $a_0=0,a_1=16,a_2=-2,a_3=142$, we conclude that

$\begin{cases} 0=a_0=c_1+c_2+c_3+c_4\\ 16=a_1=3c_1-4c_2+2c_3-2c_4\\ -2=a_2=9c_1+16c_2+4c_3+4c_4\\ 142=a_3=27c_1-64c_2+8c_3-8c_4 \end{cases}$

$\begin{cases} c_1+c_2+c_3+c_4=0\\ 3c_1-4c_2+2c_3-2c_4=16\\ 5c_1+12c_2=-2\\ 15c_1-48c_2=78 \end{cases}$

$\begin{cases} 1+c_3+c_4=0\\ 10+2c_3-2c_4=16\\ c_2=-1\\ c_1=2 \end{cases}$

$\begin{cases} c_4=-2\\ c_3=1\\ c_2=-1\\ c_1=2 \end{cases}$

It follows that the final solusion is

$a_n=2\cdot3^n-(-4)^n+2^n-2\cdot (-2)^n.$