Question

Let for x∈R, S₀(x)=x,Sk(x)=Ckx+k\(∫^x_0S_{k−1}\)(t)dt, where C₀=1,C_k≡1−\(∫^1_0S_{k−1}\)(x)dx,k==1,2,3,..... Then s₂(3)+6c₃ is equal to___.

Answer 1

Correct Answer: 18

Given Information:

\(S_0(x) = x\)

\(S_k(x) = C_k x + k \int_0^x S_{k-1}(t) \, dt\)

\(C_0 = 1\)

\(C_k = - \int_0^1 S_{k-1}(x) \, dx\) for \(k = 1, 2, \dots\)

Calculations of \(C_k\):

1. \(C_1\):

\(C_1 = - \int_0^1 S_0(x) \, dx = - \int_0^1 x \, dx = - \left[ \frac{x^2}{2} \right]_0^1 = - \frac{1}{2}\)

2. \(S_1(x)\):

\(S_1(x) = C_1 x + 1 \int_0^x S_0(t) \, dt = -\frac{1}{2} x + \int_0^x t \, dt = -\frac{1}{2} x + \frac{x^2}{2}\)

3. \(C_2\):

\(C_2 = - \int_0^1 S_1(x) \, dx = - \int_0^1 \left( -\frac{1}{2} x + \frac{1}{2} x^2 \right) dx = - \left[ -\frac{x^2}{4} + \frac{x^3}{6} \right]_0^1 = - \left( -\frac{1}{4} + \frac{1}{6} \right) = - \left( -\frac{1}{12} \right) = \frac{1}{12}\)

4. \(S_2(x)\):

\(S_2(x) = C_2 x + 2 \int_0^x S_1(t) \, dt = \frac{1}{12} x + 2 \int_0^x \left( -\frac{1}{2} t + \frac{1}{2} t^2 \right) dt = \frac{1}{12} x + 2 \left[ -\frac{t^2}{4} + \frac{t^3}{6} \right]_0^x = \frac{1}{12} x + 2 \left( -\frac{x^2}{4} + \frac{x^3}{6} \right) = \frac{1}{12} x - \frac{x^2}{2} + \frac{x^3}{3}\)

5. \(C_3\):

\(C_3 = - \int_0^1 S_2(x) \, dx = - \int_0^1 \left( \frac{1}{12} x - \frac{x^2}{2} + \frac{x^3}{3} \right) dx = - \left[ \frac{x^2}{24} - \frac{x^3}{6} + \frac{x^4}{12} \right]_0^1 = - \left( \frac{1}{24} - \frac{1}{6} + \frac{1}{12} \right) = - \left( \frac{1 - 4 + 2}{24} \right) = - \left( -\frac{1}{24} \right) = \frac{1}{24}\)

6. \(S_2(3)\):

\(S_2(3) = \frac{1}{12} (3) - \frac{(3)^2}{2} + \frac{(3)^3}{3} = \frac{1}{4} - \frac{9}{2} + 9 = \frac{1 - 18 + 36}{4} = \frac{19}{4}\)

7. \(S_2(3) + 6C_3\):

\(S_2(3) + 6C_3 = \frac{19}{4} + 6 \left( \frac{1}{24} \right) = \frac{19}{4} + \frac{1}{4} = \frac{20}{4} = 5\)

Correct Final Answer:

\( S_2(3) + 6C_3 = 5 \)