Show that $a_n=3(-2{)}^n-7(3{)}^n$ is a solution of the relation $a_n=a_{n-1}+6a_{n-2}.$

Example of verifying solution:

$d_k=4d_{k-1}-4d_{k-2},$ verify solution: $d_k=-1(2{)}^k+3k(2{)}^k$

$d_k=4d_{k-1}-4d_{k-2}$

$(1)$ Use the above to find $d_{k-1}$ and $d_{k-2}$ then $d_{k-1}$ substitute into original eq$.$

$d_k=4[-1(2{)}^{k-1}+3(k-1)(2{)}^{k-1}]-4[-1(2{)}^{k-2}+3(k-2)(2{)}^{k-2}]$

$(2)$ Use $x^{a-b}=\frac{x^a}{x^b}$ to rewrite

$d_k=4[-1\frac{2^k}{2^{\text{'}}}+3(k-1)\frac{2^k}{2^{\text{'}}}]-4[-1\frac{2^k}{2^2}+3(k-2)\frac{2^k}{2^2}]$

$(3)$ continue to carefully simplify

$d_k=-\frac{4}{2}*2^k+\frac{12}{2}(k-1)2^k+\frac{4}{4}{2}^k-\frac{12}{4}(k-2)2^k$

$=-2*2^k+6*2^k(k-1)+2^k-3*2^k(k-2)$

$=-2*2^k+6k*2^k-6*2^k+2^k-3k*2^k+6*2^k$

$=-1*2^k+3k*2^k$

This should matoh your solution for $d_k.$ It does! yay!