(1 point) Find the solution to the following recurrence: a_(n)=-8a_(n-1)-12a_(n-2)+441n for ngeqslant 2 with initial conditions a_(n)=

To solve the given recurrence relation, we can use the method of solving linear homogeneous recurrence relations with constant coefficients.<br /><br />The given recurrence relation is:<br />$a_{n}=-8a_{n-1}-12a_{n-2}+441n$ for $n\geqslant 2$<br /><br />We can rewrite this as:<br />$a_{n}+8a_{n-1}+12a_{n-2}=441n$<br /><br />Now, let's assume a solution of the form $a_{n}=r^{n}$, where $r$ is a constant.<br /><br />Substituting this into the recurrence relation, we get:<br />$r^{n}+8r^{n-1}+12r^{n-2}=441n$<br /><br />Dividing both sides by $r^{n-2}$, we get:<br />$r^{2}+8r+12=\frac{441n}{r^{2}}$<br /><br />Now, we can solve this quadratic equation to find the values of $r$.<br /><br />The quadratic equation is:<br />$r^{2}+8r+12=0$<br /><br />Solving this equation, we get:<br />$r_{1}=-2$ and $r_{2}=-6$<br /><br />Now, we can write the general solution of the recurrence relation as:<br />$a_{n}=c_{1}(-2)^{n}+c_{2}(-6)^{n}$<br /><br />To find the values of $c_{1}$ and $c_{2}$, we can use the initial conditions.<br /><br />Let's assume $a_{0}=0$ and $a_{1}=0$ as initial conditions.<br /><br />Substituting these values into the general solution, we get:<br />$0=c_{1}(-2)^{0}+c_{2}(-6)^{0}$<br />$0=c_{1}+c_{2}$<br /><br />Solving this equation, we get:<br />$c_{1}=-c_{2}$<br /><br />Now, we can substitute this value of $c_{1}$ into the general solution:<br />$a_{n}=-c_{2}(-2)^{n}+c_{2}(-6)^{n}$<br /><br />Simplifying this expression, we get:<br />$a_{n}=c_{2}((-6)^{n}-(-2)^{n})$<br /><br />To find the value of $c_{2}$, we can use the initial condition $a_{1}=0$.<br /><br />Substituting $n=1$ into the expression for $a_{n}$, we get:<br />$0=c_{2}((-6)^{1}-(-2)^{1})$<br />$0=c_{2}(-6+2)$<br />$0=-4c_{2}$<br /><br />Solving this equation, we get:<br />$c_{2}=0$<br /><br />Therefore, the solution to the given recurrence relation is:<br />$a_{n}=0$<br /><br />So, the final answer is $a_{n}=0$.