Let for x ∈ ℝ, S _ 0 (x) = x, S _ k (x) = C _ k x + k ∫ _ 0 ^ x S _...

Given,

${S}_{0}\left(x\right) = x, {S}_{k}$\left(x\right) = {C}_{k}x + k{∫}_{0}^{x}{S}_{k - 1}\left(t\right)dt$</p><p>Now for${S}_{0}\left(x\right) = x, {C}_{0} = 1$</p><p>Now solving,${C}_{k} = 1 - {∫}_{0}^{1}{S}_{k - 1}\left(x\right)dx$for$k = 1$we get,</p><p>${C}_{1} = 1 - {∫}_{0}^{1}xdx = \dfrac{1}{2}$</p><p>Now putting$k = 1$in${S}_{k}\left(x\right) = {C}_{k}x + k{∫}_{0}^{x}{S}_{k - 1}$\left(t\right)dt$ we get,

$⇒ {S}_{1}\left(x\right) = \dfrac{x}{2} + 1 \cdot {∫}_{0}^{x}tdt = \dfrac{x}{2} + \dfrac{{x}^{2}}{2}$

Now for $k = 2$ we get,

${C}_{2} = 1 - {∫}_{0}^{1}\left(\dfrac{x}{2} + \dfrac{{x}^{2}}{2}\right)dx = \dfrac{7}{12}$

And ${S}_{2}\left(x\right) = \dfrac{7}{12}x + 2{∫}_{0}^{x}\left(\dfrac{t}{2} + \dfrac{{t}^{2}}{2}\right)dt$

$⇒ {S}_{2}\left(x\right) = \dfrac{7x}{12} + \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3}$

Now taking $k = 3$ we get,

${C}_{3} = 1 - {∫}_{0}^{1}\left(\dfrac{7x}{12} + \dfrac{{x}^{2}}{2} + \dfrac{{x}^{3}}{3}\right)dx = \dfrac{11}{24}$

Now finding the value of${S}_{2}\left(3\right) + 6 \cdot {C}_{3}$ we get,

${S}_{2}\left(3\right) + 6 \cdot {C}_{3} = \dfrac{7 \times 3}{12} + \dfrac{{3}^{2}}{2} + \dfrac{{3}^{3}}{3} + 6 \times \dfrac{11}{24}$

$⇒ {S}_{2}\left(3\right) + 6 \cdot {C}_{3} = \dfrac{61}{4} + \dfrac{11}{4} = 18$