GivenU-n-0-1-x-n-1-x-dx-n-N-show-that-U-n-2-n-2-n-n-1-2n-3-

Question Number 98271 by Ar Brandon last updated on 12/Jun/20

G {ivenU}_{n} = \int_{0}^{1} x^{n} \sqrt{1 - x} dx n \in N, show that

U_{n} = \frac{2^{n + 2} n! (n + 1)}{(2 n + 3)!}

Commented by Ar Brandon last updated on 12/Jun/20

Thanks �� But, it really isn't clear to me. Any explanations please ?��

Commented by PRITHWISH SEN 2 last updated on 12/Jun/20

then U_{n} = \frac{2 n}{2 n + 3} . \frac{2 (n - 1)}{2 n + 1} \dots \dots \frac{4}{7} . \frac{2}{5} . \frac{2}{3}

= \frac{2^{n} {n (n - 1) (n - 2) (n - 3) \dots \dots 2.1} . 2}{(2 n + 3) (2 n + 1) \dots \dots \dots 7.5.3}

= \frac{2^{n + 1} . n!}{(2 n + 3)!} \times 2^{n + 1} {(n + 1) n (n - 1) \dots \dots . . 2.1}

= \frac{2^{2 n + 2} n! (n + 1)!}{(2 n + 3)!} I got this please check

Commented by Ar Brandon last updated on 12/Jun/20

I^{'} m stucked here . Can anyone help me out, please ?

Commented by Ar Brandon last updated on 12/Jun/20

Commented by PRITHWISH SEN 2 last updated on 20/Jul/20

I did not do anything I just expanded your expression

U_{n} = \frac{2 n}{2 n + 3} U_{n - 1} then U_{n - 1} = \frac{2 (n - 1)}{2 (n - 1) + 3} U_{n - 2}

\Rightarrow U_{n - 1} = \frac{2 (n - 1)}{2 n + 1} U_{n - 2} and likewise you can get

by putting n = n - 2, n - 3 and so on

now for U_{1} = \frac{2}{5} U_{0} {andU}_{0} = \int_{0}^{1} x^{0} \sqrt{1 - x} dx = \frac{2}{3}

I think now it makes sense

Answered by smridha last updated on 12/Jun/20

l e t x = \sin^{2} θ s o

2 \int_{0}^{\frac{π}{2}} {s i n}^{(2 n + 1)} θ {c o s}^{2} θ d θ

= \frac{Γ (n + 1) . Γ (\frac{3}{2})}{Γ (n + \frac{5}{2})} = \frac{n! \sqrt{π}}{(2 n + 3) . Γ (n + \frac{3}{2})}

I t h i n k n o w y o u c a n e a s y l y

g e t y o u r a n s .

Commented by PRITHWISH SEN 2 last updated on 13/Jun/20

wow! excellent sir .

Commented by smridha last updated on 13/Jun/20

t h a n k s . .

Commented by PRITHWISH SEN 2 last updated on 13/Jun/20

welcome

Commented by Ar Brandon last updated on 13/Jun/20

Hum, thank you ��

Answered by Dwaipayan Shikari last updated on 01/Nov/20

\int_{0}^{1} x^{n} {(1 - x)}^{\frac{1}{2}} d x

= \int_{0}^{1} x^{n + 1 - 1} x^{\frac{3}{2} - 1} d x

= β (n + 1, \frac{3}{2}) = \frac{Γ (n + 1) Γ (\frac{3}{2})}{Γ (n + \frac{5}{2})}