The value of k N for which the integral I_n= _1^0(1-x^k)^ndx,n N, satisfies 147I_20=148I_2…

Question

The value of $k\in N$ for which the integral ${I}_{n}={\int }_{1}^{0}{(1-{x}^{k})}^{n}dx,n\in N$, satisfies $147{I}_{20}=148{I}_{21}$ is

Accepted Answer

$\begin{gathered} I(21)=\int _{0}^{1}{(1-{x}^{k})}^{21}dx \ =\int _{0}^{1}(1-{x}^{k}){(1-{x}^{k})}^{20}dx \ =\int _{0}^{1}{(1-{x}^{k})}^{20}dx-{\int }_{0}{x}^{k}{(1-{x}^{k})}^{20}dx \ I(21)=I(20)-\int _{0}^{1}{x}^{k}{(1-{x}^{k})}^{20}dx \end{gathered}$

$I(21)=I(20)-\lfloor \frac{{(1-{x}^{k})}^{21}}{-21k}x-\int _{0}^{1}\frac{(1-{x}^{k}{)}^{21}}{-21k}dxfloor $

$\begin{gathered} I(21)=I(20)-\frac{1}{21k}I(20) \ \Rightarrow [I(21)](21k+1)=21KI(20) \ \Rightarrow 21K=147\Rightarrow K=7 \end{gathered}$

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