x in R , S_0( x )= x,S _{ k }( x )= C _{ k } | vedclass

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Question

$S _{ k }( x )=C_k x + k \int_0^x S_{k-1}(t) d t$

Put $k=2$ and $x=3$

$S _2(3)= C _2(3)+2 \int_0^3 S _1( t ) dt$

Also,

$S_1(x)=C_1(x)+\int

Vedclass · Accepted Answer

$18$

Vedclass · Answer

$S _{ k }( x )=C_k x + k \int_0^x S_{k-1}(t) d t$

Put $k=2$ and $x=3$

$S _2(3)= C _2(3)+2 \int_0^3 S _1( t ) dt$

Also,

$S_1(x)=C_1(x)+\int_0^x S_0(t) d t$

$=C_1 x+\frac{x^2}{2}$

$S_2(3)=3 C_2+2 \int_0^3\left(C_1 t+\frac{t^2}{2}\right) d t$

$=3 C_2+9 C_1+9$

Also,

$C _1=1-\int_0^1 S _0( x ) dx =\frac{1}{2}$

$C _2=1-\int_0^1 S _1( x ) dx =0$

$C _3=1-\int_0^1 S _2( x ) dx$

$=1-\int_0^1\left( C _2 x + C _1 x ^2+\frac{ x ^3}{3}\right) d x$

$=\frac{3}{4}$

$S _2( x )= C _2 x +2 \int_0^{ x } S _1( t ) dt$

$= C _2 x + C _1 x ^2+\frac{ x ^3}{3}$

$\Rightarrow S _2(3)+6 C _3=6 C _3+3 C _2+9 C _1+9$

$=18$

$x \in R , S_0( x )= x$,$S _{ k }( x )= C _{ k } x + k \int _0^{ x } S _{ k -1}(t) d t$,માટે,ધારોકે $C _0=1, C _{ k }=1-\int_0^1 S _{ k -1}( x ) dx , k =1,2,3 \ldots$. જ્યાં $S _2(3)+6 C _3$ તો $=...........$.

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