Let for $$x \in \mathbb{R}$$, $$S_0(x) = x$$, $$S_k(x) = C_k x + k\int_0^x S_{k-1}(t)dt$$ where $$C_0 = 1$$, $$C_k = 1 - \int_0^1 S_{k-1}(x)dx$$, $$k = 1, 2, 3, \ldots$$ Then $$S_2(3) + 6C_3$$ is equal to _____.

Given: $$S_0(x) = x$$, $$S_k(x) = C_k x + k\int_0^x S_{k-1}(t)\,dt$$ where $$C_0 = 1$$ and $$C_k = 1 - \int_0^1 S_{k-1}(x)\,dx$$.

Find $$C_1$$ and $$S_1(x)$$.

$$C_1 = 1 - \int_0^1 S_0(x)\,dx = 1 - \int_0^1 x\,dx = 1 - \frac{1}{2} = \frac{1}{2}$$ $$S_1(x) = \frac{1}{2}x + 1 \cdot \int_0^x t\,dt = \frac{x}{2} + \frac{x^2}{2}$$

Find $$C_2$$ and $$S_2(x)$$.

$$C_2 = 1 - \int_0^1 \left(\frac{x}{2} + \frac{x^2}{2}\right)dx = 1 - \left[\frac{x^2}{4} + \frac{x^3}{6}\right]_0^1 = 1 - \frac{1}{4} - \frac{1}{6} = 1 - \frac{5}{12} = \frac{7}{12}$$ $$S_2(x) = \frac{7}{12}x + 2\int_0^x \left(\frac{t}{2} + \frac{t^2}{2}\right)dt = \frac{7x}{12} + 2\left[\frac{t^2}{4} + \frac{t^3}{6}\right]_0^x$$ $$= \frac{7x}{12} + \frac{x^2}{2} + \frac{x^3}{3}$$

Find $$S_2(3)$$.

$$S_2(3) = \frac{7 \times 3}{12} + \frac{9}{2} + \frac{27}{3} = \frac{7}{4} + \frac{9}{2} + 9 = \frac{7}{4} + \frac{18}{4} + \frac{36}{4} = \frac{61}{4}$$

Find $$C_3$$.

$$C_3 = 1 - \int_0^1 \left(\frac{7x}{12} + \frac{x^2}{2} + \frac{x^3}{3}\right)dx = 1 - \left[\frac{7x^2}{24} + \frac{x^3}{6} + \frac{x^4}{12}\right]_0^1$$ $$= 1 - \frac{7}{24} - \frac{4}{24} - \frac{2}{24} = 1 - \frac{13}{24} = \frac{11}{24}$$

Compute the answer.

$$S_2(3) + 6C_3 = \frac{61}{4} + 6 \times \frac{11}{24} = \frac{61}{4} + \frac{11}{4} = \frac{72}{4} = 18$$

The answer is $$\boxed{18}$$.