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Question Number 184349 by alcohol last updated on 05/Jan/23

$$I_n={\int }_0^1{(1-x^2)}^{n}dx$$$$Relate{I}_{n}and{I}_{n-1}$$$$Find{I}_{n}intermsofn$$$$hencededucethat\underset{k=0}{\sum ^n}\frac{{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)}{2k+1}=\frac{2^{2n}{(n!)}^2}{(2n+1)!}$$

Answered by mr W last updated on 05/Jan/23

$$I_n={\int }_0^1{(1-x^2)}^{n}dx$$$$={\left[x{(1-x^2)}^n\right]}_0^1+2n{\int }_0^1{x}^2{(1-x^2)}^{n-1}dx$$$$=2n{\int }_0^1(1+x^2-1){(1-x^2)}^{n-1}dx$$$$=2n{\int }_0^1{(1-x^2)}^{n-1}dx-2n{\int }_0^1{(1-x^2)}^{n}dx$$$$=2{nI}_{n-1}-2{nI}_n$$$$\Rightarrow I_n=\frac{2n}{2n+1}{I}_{n-1}$$$$I_n=\frac{2n}{2n+1}\times \frac{2n-2}{2n-1}\times ...\times \frac{2}{3}\times I_0$$$$I_0={\int }_0^{1}dx=1$$$$\Rightarrow I_n=\frac{\left(2n\right)!!}{(2n+1)!!}=\frac{{[\left(2n\right)!!]}^2}{(2n+1)!}=\frac{{[2^{n}n!]}^2}{(2n+1)!}=\frac{2^{2n}{(n!)}^2}{(2n+1)!}$$$$$$$$ontheotherside:$$$$I_n={\int }_0^1{(1-x^2)}^{n}dx$$$$={\int }_0^1\left[\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right){(-x^2)}^k\right]dx$$$$=\underset{k=0}{\sum ^n}{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)\left[{\int }_0^1{x}^{2k}dx\right]$$$$=\underset{k=0}{\sum ^n}{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right){\left[\frac{x^{2k+1}}{2k+1}\right]}_0^1$$$$=\underset{k=0}{\sum ^n}\frac{{(-1)}^k}{2k+1}\left(\begin{array}{c}n\\ k\end{array}\right)$$$$\Rightarrow \underset{k=0}{\sum ^n}\frac{{(-1)}^k}{2k+1}\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{2^{2n}{(n!)}^2}{(2n+1)!}$$

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