Let say r^((n)) = Π_(k=0) ^(n−1) (r−k) and r^((0)) =1 With n∈N and r∈R... 1. Show that (n−1−r)^((n)) = (−1)^((n)) (r)^((n)) 2. If m≤n, show that (r^((n)) /r^((m)) )=(r−m)^((n−m)) 3. Espress r^((n+m)) as w^((n)) w′^((m)) 4. Show that (2r)^((2n)) =2^(2n) r^((n)) (r−(1/2))^((n)) Can you help me... please...


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Question Number 115534 by Hassen_Timol last updated on 26/Sep/20

$Let say r^{(n)} = \underset{k = 0}{\prod^{n - 1}} (r - k) and r^{(0)} = 1$ $With n \in N and r \in R . . .$ $1 . Show that {(n - 1 - r)}^{(n)} = {(- 1)}^{(n)} {(r)}^{(n)}$ $2 . If m ⩽ n, show that \frac{r^{(n)}}{r^{(m)}} = {(r - m)}^{(n - m)}$ $3 . Espress r^{(n + m)} as w^{(n)} w^{' (m)}$ $4 . Show that {(2 r)}^{(2 n)} = 2^{2 n} r^{(n)} {(r - \frac{1}{2})}^{(n)}$ $Can you help me . . . please . . .$
Commented by Hassen_Timol last updated on 26/Sep/20

$Could you help me . . . I {don}^{'} t understand ?$
	Answered by aleks041103 last updated on 26/Sep/20

	$1 .$ ${(n - 1 - r)}^{(n)} = \underset{k = 0}{\prod^{n - 1}} (n - 1 - r - k) =$ $= \underset{k = 0}{\prod^{n - 1}} ((n - 1 - k) - r)$ $l e t i = n - 1 - k, s o i f k \in [0, n - 1] t h e n$ $i \in [n - 1, 0] \equiv [0, n - 1]$ $\underset{k = 0}{\prod^{n - 1}} ((n - 1 - k) - r) = \underset{i = 0}{\prod^{n - 1}} (i - r) =$ $= \underset{i = 0}{\prod^{n - 1}} (r - i) (- 1) = \underset{i = 0}{\prod^{n - 1}} (r - i) \underset{i = 0}{\prod^{n - 1}} (- 1) =$ $= r^{(n)} {(- 1)}^{n}$ $\Rightarrow {(n - 1 - r)}^{(n)} = {(- 1)}^{n} r^{(n)}$
	Commented by Hassen_Timol last updated on 26/Sep/20
	Thank you so much, it's very nice ! thanks a lot... if you know how to do for the other questions, it would be even more nice ! But once again, thanks a lot
	Commented by aleks041103 last updated on 27/Sep/20

	$I^{'} l l a n s w e r 3 a n d 4 l a t e r$
	Answered by aleks041103 last updated on 26/Sep/20

	$2 .$ $\frac{r^{(n)}}{r^{(m)}} = \frac{\underset{k = 0}{\prod^{n - 1}} (r - k)}{\underset{k = 0}{\prod^{m - 1}} (r - k)} = \frac{\underset{k = 0}{\prod^{m - 1}} (r - k) \underset{k = m}{\prod^{n - 1}} (r - k)}{\underset{k = 0}{\prod^{m - 1}} (r - k)}$ $= \underset{k = m}{\prod^{n - 1}} (r - k)$ $l e t k = m + i, s o i f k \in [m, n - 1] t h e n$ $i \in [0, (n - m) - 1]$ $\frac{r^{(n)}}{r^{(m)}} = \underset{i = 0}{\prod^{(n - m) - 1}} (r - (m + i)) =$ $= \underset{i = 0}{\prod^{(n - m) - 1}} ((r - m) - i) = {(r - m)}^{(n - m)}$ $\Rightarrow \frac{r^{(n)}}{r^{(m)}} = {(r - m)}^{(n - m)}$
	Commented by Hassen_Timol last updated on 26/Sep/20
	Thank you very much !
	Answered by aleks041103 last updated on 27/Sep/20

	$3 .$ $r^{(n + m)} = \underset{k = 0}{\prod^{n + m - 1}} (r - k) =$ $= \underset{k = 0}{\prod^{n - 1}} (r - k) \underset{k = n}{\prod^{n + m - 1}} (r - k) =$ $= r^{(n)} \underset{k = n}{\prod^{n + m - 1}} (r - k)$ $l e t k = n + i s o$ $\underset{k = n}{\prod^{n + m - 1}} (r - k) = \underset{i = 0}{\prod^{m - 1}} (r - n - i) = {(r - n)}^{(m)}$ $\Rightarrow r^{(n + m)} = r^{(n)} {(r - n)}^{(m)}$ $B y a n a l o g y$ $r^{(n + m)} = r^{(m)} {(r - m)}^{(n)}$ $S o$ $r^{(n + m)} = {(r - m)}^{(n)} r^{(m)} = r^{(n)} {(r - n)}^{(m)}$
	Answered by aleks041103 last updated on 27/Sep/20

	$4 .$ ${(2 r)}^{(2 n)} = \underset{k = 0}{\prod^{2 n - 1}} (2 r - k) = \underset{k = 0}{\prod^{2 n - 1}} (r - \frac{k}{2}) 2 =$ $= 2^{2 n} \underset{k = 0}{\prod^{2 n - 1}} (r - \frac{k}{2})$ $\underset{k = 0}{\prod^{2 n - 1}} (r - \frac{k}{2}) =$ $= \underset{k = 0; e v e n k}{\prod^{2 n - 1}} (r - \frac{k}{2}) \underset{k = 0; o d d k}{\prod^{2 n - 1}} (r - \frac{k}{2}) =$ $= \underset{k = 0}{\prod^{n - 1}} (r - \frac{2 k}{2}) \underset{k = 0}{\prod^{n - 1}} (r - \frac{2 k + 1}{2}) =$ $= \underset{k = 0}{\prod^{n - 1}} (r - k) \underset{k = 0}{\prod^{n - 1}} (r - \frac{1}{2} - k)$ $= r^{(n)} {(r - \frac{1}{2})}^{(n)}$ $\Rightarrow {(2 r)}^{(2 n)} = 2^{2 n} r^{(n)} {(r - \frac{1}{2})}^{(n)}$
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