Let for x ∈ ℝ ,   S 0 x = x ,   S k x = C k x + k ∫ 0 x S k - 1 t d t where C 0…

Given,

$S_0\left(x\right)=x, S_k\left(x\right)=C_{k}x+k{\int }_0^x{S}_{k-1}\left(t\right)dt$ 

Now for $S_0\left(x\right)=x,C_0=1$

Now solving, $C_k=1-{\int }_0^1{S}_{k-1}\left(x\right)dx$ for $k=1$ we get,

$C_1=1-{\int }_0^{1}xdx=\frac{1}{2}$

Now putting $k=1$ in $S_k\left(x\right)=C_{k}x+k{\int }_0^x{S}_{k-1}\left(t\right)dt$ we get,

$\Rightarrow S_1\left(x\right)=\frac{x}{2}+1·{\int }_0^{x}tdt=\frac{x}{2}+\frac{x^2}{2}$

Now for $k=2$ we get,

$C_2=1-{\int }_0^1\left(\frac{x}{2}+\frac{x^2}{2}\right)dx=\frac{7}{12}$

And $S_2\left(x\right)=\frac{7}{12}x+2{\int }_0^x\left(\frac{t}{2}+\frac{t^2}{2}\right)dt$

$\Rightarrow S_2\left(x\right)=\frac{7x}{12}+\frac{x^2}{2}+\frac{x^3}{3}$

Now taking $k=3$ we get,

$C_3=1-{\int }_0^1\left(\frac{7x}{12}+\frac{x^2}{2}+\frac{x^3}{3}\right)dx=\frac{11}{24}$

Now finding the value of$S_2\left(3\right)+6·C_3$ we get,

$S_2\left(3\right)+6·C_3=\frac{7\times 3}{12}+\frac{3^2}{2}+\frac{3^3}{3}+6\times \frac{11}{24}$

$\Rightarrow S_2\left(3\right)+6·C_3=\frac{61}{4}+\frac{11}{4}=18$