Recurring Styles 10 - Simultaneous System of Recurrences!

We have $D_{n+1} = 8L_n + 39n - 9 \quad ...(1)$

We also have $8D_n + 72n = 8L_n + 48n \quad ...(2)$

By $(1) - (2)$ , we obtain: $D_{n+1} = 8D_n + 63n - 9 \quad ...(3)$

Using the above equation, we can observe that:

$\begin{cases} D_0 = 1 = 8^0 - 9(0) \\ D_1 = -1 = 8^1 - 9(1) \\D_2 = 46 = 8^2 - 9(2) \\ D_3 = 485 = 8^3 - 9(3) \end{cases}$

Let's show by induction that $D_n = 8^n - 9(n) \quad ...(4)$

Now, $(4)$ is true for $n=0,1,2,3$ as seen by the examples.

Suppose $D_n = 8^n - 9n$

$\Rightarrow D_{n+1} = 8D_n + 63n - 9 = 8(8^n-9n)+63n-9 = 8^{n+1} - 9(n+1)$ .

Hence, by induction, we proved that $\boxed{D_n = 8^n - 9n}$

Now, by $(2)$ , we get $\boxed{L_n = 8^n - 6n}$

We have $K_{3n} = L_n + 6n - 33n = 8^n - 33n = 2^{3n} - 11(3n)$

Substituting $3n$ by $n$ we get $\boxed{K_n = 2^n - 11n}$

So $D_6 - K_{15} - L_5 = (8^6-9 \cdot 6)-(8^5-6 \cdot 5)-(2^{15}-11 \cdot 15) = \boxed{196749}$