Let $$\{a_k\}$$ and $$\{b_k\}$$, $$k \in \mathbb{N}$$, be two G.P.s with common ratio $$r_1$$ and $$r_2$$ respectively such that $$a_1 = b_1 = 4$$ and $$r_1 < r_2$$. Let $$c_k = a_k + b_k$$, $$k \in \mathbb{N}$$. If $$c_2 = 5$$ and $$c_3 = \frac{13}{4}$$ then $$\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$$ is equal to

Find $$\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$$ where $$c_k = a_k + b_k$$ with given conditions.

$$c_2 = 4r_1 + 4r_2 = 5 \Rightarrow r_1 + r_2 = \frac{5}{4}$$

$$c_3 = 4r_1^2 + 4r_2^2 = \frac{13}{4} \Rightarrow r_1^2 + r_2^2 = \frac{13}{16}$$

$$(r_1 + r_2)^2 = r_1^2 + 2r_1 r_2 + r_2^2$$

$$\frac{25}{16} = \frac{13}{16} + 2r_1 r_2 \Rightarrow r_1 r_2 = \frac{3}{8}$$

$$r_1, r_2$$ are roots of $$t^2 - \frac{5}{4}t + \frac{3}{8} = 0$$, i.e., $$8t^2 - 10t + 3 = 0$$.

$$t = \frac{10 \pm 2}{16}$$, so $$r_1 = \frac{1}{2}$$, $$r_2 = \frac{3}{4}$$ (since $$r_1 < r_2$$).

$$\sum_{k=1}^{\infty} c_k = \frac{4}{1 - r_1} + \frac{4}{1 - r_2} = \frac{4}{1/2} + \frac{4}{1/4} = 8 + 16 = 24$$

$$a_6 = 4 \cdot \left(\frac{1}{2}\right)^5 = \frac{4}{32} = \frac{1}{8}$$

$$b_4 = 4 \cdot \left(\frac{3}{4}\right)^3 = 4 \cdot \frac{27}{64} = \frac{27}{16}$$

$$12a_6 + 8b_4 = 12 \cdot \frac{1}{8} + 8 \cdot \frac{27}{16} = \frac{3}{2} + \frac{27}{2} = 15$$

$$\sum c_k - (12a_6 + 8b_4) = 24 - 15 = 9$$

The answer is $$\boxed{9}$$.