Answer to Question #238825 in Discrete Mathematics for Suddu

Question

Solve the recurrence relation an=6 an-2-12 an-2 + 8 an-2 + n2 2n

Answer 1

Let us solve the recurrence relation $a_n=6 a_{n-2}-12a_{n-2} + 8a_{n-2} + n^2+ 2n,$ which is equivalent to $a_n-2a_{n-2}= n^2+ 2n.$ The characteristic equation $k^2-2=0$ of the homogeneous recurence relation $a_n-2a_{n-2}=0$ has the roots $k_1=\sqrt{2},\ k_2=-\sqrt{2}.$

It follows that the solution of $a_n-2a_{n-2}= n^2+ 2n$ is of the form $a_n=C_1(\sqrt{2})^n+C_2(-\sqrt{2})^n+b_p(n),$ where $b_p(n)=an^2+bn+c.$ Then $an^2+bn+c-2(a(n-2)^2+b(n-2)+c)=n^2+2n,$ and hence $an^2+bn+c-2(an^2-4an+4a+bn-2b+c)=n^2+2n.$ It follows that $-an^2+(8a-b)n-8a+4b-c=n^2+2n,$ and thus $a=-1,\ 8a-b=2,\ -8a+4b-c=0.$ Therefore, $a=-1,\ b=-10,\ c= -32.$

We conclude that the solution is the following:

$a_n=C_1(\sqrt{2})^n+C_2(-\sqrt{2})^n-n^2-10n-32.$