So I have the recurrence relation
$$T_{n+1}=20T_n-8^n48.$$ For $T_0 = 6$, the first terms are $72$, $1056$, $18048$.
I've seen a few worked examples for simplifying other recurrence series, but I'm not entirely sure how to work around that $8^n$.
I'd appreciate any help with finding the single variable equation for this recurrence relation. Thanks.
$$\begin{align}T_1&=20\cdot T_0 - 48\cdot 8^0\\ T_2&= 20^2T_0 - 48 (20\cdot 8^0 + 8^1)\\ T_3 &= 20^3T_0 - 48(20^2\cdot 8^0 + 20\cdot 8^1 + 8^2) \end{align}$$
Seeing the pattern:
$$\begin{align}T_{n}&=20^nT_0 -48\sum_{k=0}^{n-1}20^{n-1-k}8^k\\ &=20^nT_0-48 \frac{20^{n}-8^{n}}{20-8}\\ &=6\cdot 20^n - 4 (20^n-8^n)\\ &=2\cdot 20^n + 4\cdot 8^n \end{align}$$
This shows that $T_n$ satisfies a linear recurrence:
$$T_{n+1}=28T_n - 160T_{n-1}$$
HINT:
Let $T_m=U_m\cdot8^m$
$\implies8^{n+1}U_{n+1}=20\times8^n\cdot U_n-48\cdot8^n$
$\implies2U_{n+1}=5\cdot U_n-12$
Let $U_m=a\cdot V_m+b$
$\implies2(a\cdot V_{n+1}+b)=5(a\cdot V_n+b)-12$
$2a\cdot V_{n+1}=5a\cdot V_n+3b-12$
Set $3b-12=0\iff b=4,a\ne0$ to get $V_{n+1}=\dfrac52V_n=\cdot=\left(\dfrac52\right)^rV_{n-r+1}$ for integer $r\le n+1$
