Let for x R,S_0(x)=x,S_k(x)=C_kx+k _x^0S_k-1(t)dt, where C_0=1,C_k=1- _1^0S_k-1(x)dx,k=1,2…

Question

Let for $x\in R,{S}_{0}(x)=x,{S}_{k}(x)={C}_{k}x+k{\int }_{x}^{0}{S}_{k-1}(t)dt$, where

${C}_{0}=1,{C}_{k}=1-{\int }_{1}^{0}{S}_{k-1}(x)dx,k=1,2,3,\ldots$ Then ${S}_{2}(3)+6{C}_{3}$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 18

Accepted Answer

Given,

${S}_{k}(x)={C}_{k}x+k{\int }_{x}^{0}{S}_{k-1}(t)dt$

Put $k=2$ and $x=3$

${S}_{2}(3)={C}_{2}(3)+2{\int }_{3}^{0}{S}_{1}(t)dt$ .........(i)

Also,

$\begin{gathered} {S}_{1}(x)={C}_{1}(x)+{\int }_{x}^{0}{S}_{0}(t)dt \ ={C}_{1}x+\frac{{x}^{2}}{2} \ {S}_{2}(3)=3{C}_{2}+2{\int }_{3}^{0}({C}_{1}t+\frac{{t}^{2}}{2})dt \ =3{C}_{2}+9{C}_{1}+9 \end{gathered}$

Also,

$\begin{gathered} {C}_{1}=1-{\int }_{1}^{0}{S}_{0}(x)dx=\frac{1}{2} \ {C}_{2}=1-{\int }_{1}^{0}{S}_{1}(x)dx=0 \ {C}_{3}=1-{\int }_{1}^{0}{S}_{2}(x)dx \ =1-{\int }_{1}^{0}({C}_{2}x+{C}_{1}{x}^{2}+\frac{{x}^{3}}{3})dx \ =\frac{3}{4} \end{gathered}$

$\begin{gathered} {S}_{2}(x)={C}_{2}x+2{\int }_{x}^{0}{S}_{1}(t)dt \ ={C}_{2}x+{C}_{1}{x}^{2}+\frac{{x}^{3}}{3} \ ={S}_{2}(3)+6{C}_{3}=6{C}_{3}+3{C}_{2}+9{C}_{1}+9 \ =18 \end{gathered}$

Solutions · 1