Let-say-r-n-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-

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

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...

Let say r^{(n)} = \underset{k = 0}{\prod^{n - 1}} (r - k) and r^{(0)} = 1

With n \in N and r \in R \dots

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 \dots please \dots

Commented by Hassen_Timol last updated on 26/Sep/20

Could you help me... I don′t understand?

Could you help me \dots I {don}^{'} t understand ?

Answered by aleks041103 last updated on 26/Sep/20

1. (n−1−r)^((n)) =Π_(k=0) ^(n−1) (n−1−r−k)= =Π_(k=0) ^(n−1) ((n−1−k)−r) let i=n−1−k, so if k∈[0,n−1] then i∈[n−1,0]≡[0,n−1] Π_(k=0) ^(n−1) ((n−1−k)−r)=Π_(i=0) ^(n−1) (i−r)= =Π_(i=0) ^(n−1) (r−i)(−1)=Π_(i=0) ^(n−1) (r−i)Π_(i=0) ^(n−1) (−1)= =r^((n)) (−1)^n ⇒(n−1−r)^((n)) =(−1)^n r^((n))

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. (r^((n)) /r^((m)) )=((Π_(k=0) ^(n−1) (r−k))/(Π_(k=0) ^(m−1) (r−k)))=((Π_(k=0) ^(m−1) (r−k)Π_(k=m) ^(n−1) (r−k))/(Π_(k=0) ^(m−1) (r−k))) =Π_(k=m) ^(n−1) (r−k) let k=m+i, so if k∈[m,n−1] then i∈[0,(n−m)−1] (r^((n)) /r^((m)) )=Π_(i=0) ^((n−m)−1) (r−(m+i))= =Π_(i=0) ^((n−m)−1) ((r−m)−i)=(r−m)^((n−m)) ⇒(r^((n)) /r^((m)) )=(r−m)^((n−m))

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)) =Π_(k=0) ^(n+m−1) (r−k)= =Π_(k=0) ^(n−1) (r−k)Π_(k=n) ^(n+m−1) (r−k)= =r^((n)) Π_(k=n) ^(n+m−1) (r−k) let k=n+i so Π_(k=n) ^(n+m−1) (r−k)=Π_(i=0) ^(m−1) (r−n−i)=(r−n)^((m)) ⇒r^((n+m)) =r^((n)) (r−n)^((m)) By analogy r^((n+m)) =r^((m)) (r−m)^((n)) So r^((n+m)) =(r−m)^((n)) r^((m)) =r^((n)) (r−n)^((m))

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. (2r)^((2n)) =Π_(k=0) ^(2n−1) (2r−k)=Π_(k=0) ^(2n−1) (r−(k/2))2= =2^(2n) Π_(k=0) ^(2n−1) (r−(k/2)) Π_(k=0) ^(2n−1) (r−(k/2))= =Π_(k=0;even k) ^(2n−1) (r−(k/2))Π_(k=0;odd k) ^(2n−1) (r−(k/2))= =Π_(k=0) ^(n−1) (r−((2k)/2))Π_(k=0) ^(n−1) (r−((2k+1)/2))= =Π_(k=0) ^(n−1) (r−k)Π_(k=0) ^(n−1) (r−(1/2)−k) =r^((n)) (r−(1/2))^((n)) ⇒(2r)^((2n)) =2^(2n) r^((n)) (r−(1/2))^((n))

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)}

Facebook Tweet Pin

Let-say-r-n-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-

Leave a Reply Cancel reply