Performance Objective Demonstration

PO#$5$ Solve recurrence relations using appropriate techniques and verify the solution$($s$).$

Write thorough work for each prompt.

Solve the recurrence relation. $($note: you are not required to verify your solution for this problem, however if you have time it's always a good idea to check$)$

$a_n=-8a_{n-1}-16a_{n-2}$ for $n2$ with initial conditions $a_0=3,a_1=-16$

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

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

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

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

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

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

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