Let for $x \in \mathbb{R}, S_{0}(x)=x, S_{k}(x)=C_{k} x+k \int_{0}^{x} S_{k-1}(t) d t$, where

$C_{0}=1, C_{k}=1-\int_{0}^{1} S_{k-1}(x) d x, k=1,2,3, \ldots$ Then $S_{2}(3)+6 C_{3}$ is equal to ____________.

Given, <br/><br/>$S_k(x)=C_k x+k \int_0^x S_{k-1}(t) d t$ <br/><br/>Put $\mathrm{k}=2$ and $\mathrm{x}=3$ <br/><br/>$\mathrm{S}_2(3)=\mathrm{C}_2(3)+2 \int_0^3 \mathrm{~S}_1(\mathrm{t}) \mathrm{dt}$ .........(i) <br/><br/>Also, <br/><br/>$$ \begin{aligned} & S_1(x)=C_1(x)+\int_0^x S_0(t) d t \\\\ & =C_1 x+\frac{x^2}{2} \\\\ & S_2(3)=3 C_2+2 \int_0^3\left(C_1 t+\frac{t^2}{2}\right) d t \\\\ & =3 C_2+9 C_1+9 \end{aligned} $$ <br/><br/>Also, <br/><br/>$$ \begin{aligned} & \mathrm{C}_1=1-\int_0^1 \mathrm{~S}_0(\mathrm{x}) \mathrm{dx}=\frac{1}{2} \\\\ & \mathrm{C}_2=1-\int_0^1 \mathrm{~S}_1(\mathrm{x}) \mathrm{dx}=0 \\\\ & \mathrm{C}_3=1-\int_0^1 \mathrm{~S}_2(\mathrm{x}) \mathrm{dx} \\\\ & =1-\int_0^1\left(\mathrm{C}_2 \mathrm{x}+\mathrm{C}_1 \mathrm{x}^2+\frac{\mathrm{x}^3}{3}\right) \mathrm{dx} \\\\ & =\frac{3}{4} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & S_2(x)=C_2 x+2 \int_0^x S_1(t) d t \\\\ &=C_2 x+C_1 x^2+\frac{x^3}{3} \\\\ & = S_2(3)+6 C_3=6 C_3+3 C_2+9 C_1+9 \\\\ &=18 \end{aligned} $$