Let for x , S 0 ( x ) = x , S k ( x ) = C k x + k 0 x S k - 1 ( t ) ? d t , where C 0 = 1 , C k = 1 - 0 1 S k - 1 ( x ) ? d x , k = 1 , 2 , 3 , … Then S 2 ( 3 ) + 6 C 3 is equal to _______ . [2023]

Question

Let for $x \in ℝ, S_{0} (x) = x, S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t,$ where $C_{0} = 1, C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x, k = 1, 2, 3, \dots$ Then $S_{2} (3) + 6 C_{3}$ is equal to _______ . [2023]

Smart Achievers · Accepted Answer

(18)

Given, $S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t ...(i)$

Put $k$ = 2 and $x$ = 3 in (i),

$S_{2} (3) = C_{2} (3) + 2 \int_{0}^{3} S_{1} (t) d t ...(ii)$

Put $k$ = 1 in (i),

$S_{1} (x) = C_{1} x + \int_{0}^{x} S_{0} (t) d t = C_{1} x + \int_{0}^{x} t d t [∵ S_{0} (x) = x]$

$= C_{1} x + \frac{x^{2}}{2}$

Put $S_{1} (t) = C_{1} t + \frac{t^{2}}{2}$ in (ii),

$S_{2} (3) = C_{2} (3) + 2 \int_{0}^{3} (C_{1} t + \frac{t^{2}}{2}) d t$

$= 3 C_{2} + 2 {[C_{1} \frac{t^{2}}{2} + \frac{t^{3}}{6}]}_{0}^{3} = 3 C_{2} + 2 [C_{1} (\frac{9}{2}) + (\frac{27}{6})]$

$= 3 C_{2} + 9 C_{1} + 9$

Given, $C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x$ $...(iii)$

Put $k$ = 1 in (iii),

$C_{1} = 1 - \int_{0}^{1} S_{0} (x) d x = 1 - \int_{0}^{1} x d x = 1 - {(\frac{x^{2}}{2})}_{0}^{1} = \frac{1}{2}$

Put $k$ = 2 in (iii),

$C_{2} = 1 - \int_{0}^{1} S_{1} (x) d x = 1 - \int_{0}^{1} (C_{1} x + \frac{x^{2}}{2}) d x$

$= 1 - {[C_{1} \frac{x^{2}}{2} + \frac{x^{3}}{6}]}_{0}^{1} = 1 - [\frac{1}{2} \times \frac{1}{2} + \frac{1}{6}] = 1 - [\frac{1}{4} + \frac{1}{6}] = 1 - \frac{5}{12} = \frac{7}{12}$

Put $k$ = 3 in (iii), $C_{3} = 1 - \int_{0}^{1} S_{2} (x) d x$

Put $k$ = 2 in (i), $S_{2} (x) = C_{2} x + 2 \int_{0}^{x} S_{1} (t) d t = C_{2} x + C_{1} x^{2} + \frac{x^{3}}{3}$

So, $C_{3} = 1 - \int_{0}^{1} (C_{2} x + C_{1} x^{2} + \frac{x^{3}}{3}) d x$

$= 1 - {[C_{2} \frac{x^{2}}{2} + C_{1} \frac{x^{3}}{3} + \frac{x^{4}}{12}]}_{0}^{1}$

$= 1 - \frac{7}{12} \times \frac{1}{2} - \frac{1}{2} \times \frac{1}{3} - \frac{1}{12} = 1 - \frac{7}{24} - \frac{1}{6} - \frac{1}{12}$

$= \frac{24 - 7 - 4 - 2}{24} = \frac{11}{24}$

$S_{3} (3) + 6 C_{3} = 3 C_{2} + 9 C_{1} + 9 + 6 C_{3}$

$= 3 \times \frac{7}{12} + 9 \times \frac{1}{2} + 9 + 6 \times \frac{11}{24} = \frac{7}{4} + \frac{9}{2} + 9 + \frac{11}{4}$

$= \frac{7 + 18 + 36 + 11}{4} = \frac{72}{4} = 18$

Solution

Q.
Let for $x \in ℝ, S_{0} (x) = x, S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t,$ where $C_{0} = 1, C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x, k = 1, 2, 3, \dots$ Then $S_{2} (3) + 6 C_{3}$ is equal to _______ . [2023]

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Solution

Q. Let for x∈ℝ, S0(x)=x,Sk(x)=Ckx+k∫0xSk-1(t) dt, where C0=1, Ck=1-∫01Sk-1(x) dx, k=1,2,3,… Then S2(3)+6C3 is equal to _______ . [2023]

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Q.
Let for $x \in ℝ, S_{0} (x) = x, S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t,$ where $C_{0} = 1, C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x, k = 1, 2, 3, \dots$ Then $S_{2} (3) + 6 C_{3}$ is equal to _______ . [2023]