If $T\left( n \right) = 8T\left( n-1 \right) - 15T\left( n-2 \right); T\left(1\right) = 1; T\left( 2 \right) = 4$,
What's $T\left(n\right)$ ?
I use this method:
Let $c(T(n) - aT(n-1)) = T(n-1) - aT(n-2)$
from $T(n) = 8T(n-1) - 15T(n-2)$, we can get $\begin{cases}c = \frac{1}{3}\\a = 5\end{cases}$ and $\begin{cases}c = \frac{1}{5}\\a = 3\end{cases}$,
then, we get $\frac{T(n)-5T(n-1)}{T(n-1)-5T(n-2)} = 3$,
so, we reach the answer: $T(n) = \frac{3^{n-1}+5^{n-1}}{2}$.
Sorry for my poor English.