Let for $\mathrm{x}\in \mathrm{ℝ},{\mathrm{S}}_0\left(\mathrm{x}\right)=\mathrm{x},{\mathrm{S}}_{\mathrm{k}}\left(\mathrm{x}\right)={\mathrm{C}}_{\mathrm{k}}\mathrm{x}+\mathrm{k}{\int }_0^{\mathrm{x}}{\mathrm{S}}_{\mathrm{k}-1}\left(\mathrm{t}\right)\mathrm{dt},$ where ${\mathrm{C}}_0=1,{\mathrm{C}}_{\mathrm{k}}=1-{\int }_0^1{\mathrm{S}}_{\mathrm{k}-1}\left(\mathrm{x}\right)\mathrm{dx},\mathrm{k}=1,2,3......$
Then ${\mathrm{S}}_2\left(3\right)+6{\mathrm{C}}_3$ is equal to _______
 

$$\begin{gathered} {\mathrm{c}}_1=1-\underset{0}{\overset{1}{\int }}\mathrm{x}\mathrm{dx}=1-\frac{1}{2}=\frac{1}{2}, \\ {\mathrm{s}}_1\left(\mathrm{x}\right)={\mathrm{c}}_1.\mathrm{x}+\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{t}\mathrm{dt}={\mathrm{c}}_1.\mathrm{x}+{\left[\frac{{\mathrm{t}}^2}{2}\right]}_0^{\mathrm{x}}={\mathrm{c}}_1.\mathrm{x}+\frac{{\mathrm{x}}^2}{2}=\frac{\mathrm{x}}{2}+\frac{{\mathrm{x}}^2}{2}=\frac{{\mathrm{x}}^2+\mathrm{x}}{2} \end{gathered}$$
${\mathrm{s}}_2\left(\mathrm{x}\right)={\mathrm{c}}_2.\mathrm{x}+2.\underset{0}{\overset{\mathrm{x}}{\int }}\left(\frac{{\mathrm{t}}^2+\mathrm{t}}{2}\right)\mathrm{dt}={\mathrm{c}}_2\mathrm{x}+.\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^2}{2}$

${\mathrm{c}}_2=1-\underset{0}{\overset{1}{\int }}\frac{{\mathrm{x}}^2+\mathrm{x}}{2}$
$c_2=1-\frac{1}{2}(\frac{1}{3}+\frac{1}{2})=1-\frac{5}{12}=\frac{7}{12}$

${\mathrm{s}}_2\left(\mathrm{x}\right)=\frac{7}{12}\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^2}{2}\Rightarrow {\mathrm{s}}_2\left(3\right)=\frac{7}{4}+9+\frac{9}{2}=\frac{61}{4}$
${\mathrm{c}}_3=1-\underset{0}{\overset{1}{\int }}(\frac{7}{12}\mathrm{x}+\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^2}{2})\mathrm{dx}=1-(\frac{7}{12}\frac{1}{2}+\frac{1}{3}\frac{1}{4}+\frac{1}{2}\frac{1}{3})$
 

${\mathrm{c}}_3=1-(\frac{7}{24}+\frac{1}{12}+\frac{1}{6})$

$C_3=1-\left(\frac{7+2+4}{24}\right) =1-\frac{13}{24}=\frac{11}{24}$

${\mathrm{s}}_3\left(3\right)+6{\mathrm{c}}_3=\frac{61}{4}+6.\frac{11}{24}=\frac{72}{4}=18$