Question Number 41518 by maxmathsup by imad last updated on 08/Aug/18
$${calculate}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt}\:\:\:{with}\:{n}\:{integr}\:{natural} \\ $$
Answered by alex041103 last updated on 09/Aug/18
![We know that (a+b)^n =Σ_(k=0) ^n ((k),(n) )a^(n−k) b^k ⇒(1−t^2 )^n =Σ_(k=0) ^n (−1)^k ((k),(n) )t^(2k) ⇒A_n =∫_( 0) ^1 [ Σ_(k=0) ^n (−1)^k ((k),(n) )t^(2k) ]dt= =Σ_(k=0) ^n [(−1)^k ((k),(n) ) ∫_( 0) ^1 t^(2k) dt]= =Σ_(k=0) ^n (((−1)^k )/(2k+1)) ((k),(n) )](https://www.tinkutara.com/question/Q41542.png)
$${We}\:{know}\:{that} \\ $$$$\left({a}+{b}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{k}}\\{{n}}\end{pmatrix}{a}^{{n}−{k}} {b}^{{k}} \\ $$$$\Rightarrow\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \begin{pmatrix}{{k}}\\{{n}}\end{pmatrix}{t}^{\mathrm{2}{k}} \\ $$$$\Rightarrow{A}_{{n}} =\underset{\:\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \begin{pmatrix}{{k}}\\{{n}}\end{pmatrix}{t}^{\mathrm{2}{k}} \:\right]{dt}= \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\left[\left(−\mathrm{1}\right)^{{k}} \begin{pmatrix}{{k}}\\{{n}}\end{pmatrix}\:\underset{\:\:\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}{t}^{\mathrm{2}{k}} {dt}\right]= \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \:}{\mathrm{2}{k}+\mathrm{1}}\begin{pmatrix}{{k}}\\{{n}}\end{pmatrix}\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
$${excellent}… \\ $$
Answered by math khazana by abdo last updated on 09/Aug/18
![we have A_(n+1) =∫_0 ^1 (1−t^2 )^(n+1) dt =∫_0 ^1 (1−t^2 )^n (1−t^2 )dt= ∫_0 ^1 (1−t^2 )^n −∫_0 ^1 t^2 (1−t^2 )^n dt =A_n −∫_0 ^1 t^2 (1−t^2 )^n dt by parts ∫_0 ^1 t^2 (1−t^2 )^n dt =−(1/2) ∫_0 ^1 t(−2t)(1−t^2 )^n dt =−(1/2){ [ (t/(n+1))(1−t^2 )^(n+1) ]_0 ^1 −∫_0 ^1 (1/(n+1))(1−t^2 )^(n+1) dt} =(1/(2(n+1))) A_(n+1) ⇒(1−(1/(2(n+1))))A_(n+1) =A_n ⇒ ((2n+1)/(2n+2)) A_(n+1) = A_n ⇒A_(n+1) =((2n+2)/(2n+1)) A_n ⇒ Π_(k=0) ^(n−1) A_(k+1) =Π_(k=0) ^(n−1) ((2k+2)/(2k+1)) Π_(k=0) ^(n−1) A_n ⇒ A_n = Π_(k=0) ^(n−1) ((2k+2)/(2k+1)) A_(0 ) but A_0 =1 ⇒ A_n =((2.4.8.....(2n))/(1.3.5......(2n−1))) =((2^n (n!) 2.4.8.....(2n))/(1.2.3.4.5.....(2n))) =((2^(2n) (n!)^2 )/((2n)!)) .](https://www.tinkutara.com/question/Q41546.png)
$${we}\:{have}\:{A}_{{n}+\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt} \\ $$$$={A}_{{n}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt}\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} \:{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\left(−\mathrm{2}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\left[\:\:\frac{{t}}{{n}+\mathrm{1}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} {dt}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\:{A}_{{n}+\mathrm{1}} \:\Rightarrow\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\right){A}_{{n}+\mathrm{1}} ={A}_{{n}} \:\Rightarrow \\ $$$$\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\:{A}_{{n}+\mathrm{1}} =\:{A}_{{n}} \:\Rightarrow{A}_{{n}+\mathrm{1}} =\frac{\mathrm{2}{n}+\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}}\:{A}_{{n}} \:\:\Rightarrow \\ $$$$\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{A}_{{k}+\mathrm{1}} =\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{2}{k}+\mathrm{2}}{\mathrm{2}{k}+\mathrm{1}}\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{A}_{{n}} \:\Rightarrow \\ $$$${A}_{{n}} =\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\mathrm{2}{k}+\mathrm{2}}{\mathrm{2}{k}+\mathrm{1}}\:{A}_{\mathrm{0}\:} \:\:\:\:\:{but}\:{A}_{\mathrm{0}} =\mathrm{1}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{2}.\mathrm{4}.\mathrm{8}…..\left(\mathrm{2}{n}\right)}{\mathrm{1}.\mathrm{3}.\mathrm{5}……\left(\mathrm{2}{n}−\mathrm{1}\right)}\:=\frac{\mathrm{2}^{{n}} \left({n}!\right)\:\mathrm{2}.\mathrm{4}.\mathrm{8}…..\left(\mathrm{2}{n}\right)}{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}…..\left(\mathrm{2}{n}\right)} \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:. \\ $$
Commented by alex041103 last updated on 10/Aug/18
$${Perfect}!\:{Deserves}\:{a}\:{like}! \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
$${let}\:{t}={sin}\propto\:\:{dt}={cos}\propto{d}\propto \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}} \propto{cos}\propto\:{d}\propto \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}+\mathrm{1}} \propto.{sin}^{\mathrm{0}} \propto\:{d}\propto \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {cos}^{\mathrm{2}\left({n}+\mathrm{1}\right)−\mathrm{1}} .{sin}^{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \propto{d}\propto \\ $$$$\frac{\lceil\left(\mathrm{2}{n}+\mathrm{2}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\lceil\left(\mathrm{2}{n}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\:\:\lceil\left({n}\right)={gamma}\:{function} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
