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Question Number 57515 by Hassen_Timol last updated on 06/Apr/19

Commented by Hassen_Timol last updated on 06/Apr/19

$E x c u s e m e, w h a t d o e s i t m e a n ?$ $C a n y o u g i v e s o m e e x a m p l e s f o r n ?$
Commented by maxmathsup by imad last updated on 06/Apr/19
$let prove first that (x+1)^n =Σ_(k=0) ^n C_n ^k x^k with C_n ^k =((n!)/(k!(n−k)!)) by recurrence let put P_n (x) =(x+1)^n −Σ_(k=0) ^n C_n ^k x^k let prove that P_n (x)=0 P_0 (x) =0 (true) let suppose P_n (x)=0 we have P_(n+1) (x) =(x+1)^(n+1) −Σ_(k=0) ^(n+1) C_(n+1) ^k x^k =(x+1)^n (x+1)−Σ_(k=0) ^(n+1) C_(n+1) ^k x^k =(x+1){P_n (x)+Σ_(k=0) ^n C_n ^k x^k }−Σ_(k=0) ^(n+1) C_(n+1) ^k x^k but P_n (x)=0 ⇒ P_(n+1) (x)=(x+1)Σ_(k=0) ^n C_n ^k x^k −1−Σ_(k=1) ^(n+1) {C_n ^k +C_n ^(k−1) }x^k =x Σ_(k=0) ^n C_n ^k x^k +Σ_(k=0) ^n C_n ^k x^k −1 −Σ_(k=1) ^n C_n ^k x^k −Σ_(k=1) ^(n+1) C_n ^(k−1) x^k =xΣ_(k=0) ^n C_n ^k x^k −Σ_(p=0) ^n C_n ^p x^(p+1) ( changement k−1=p) =x(Σ_(k=0) ^n C_n ^k x^k −Σ_(p=0) ^n C_n ^p x^p ) =0 ⇒P_(n+1) (x)=0 so (x+1)^(n ) =Σ_(k=0) ^n C_n ^k x^k we have (x+y)^n =x^n if y=0 let suppose y≠0 ⇒ (x+y)^n =y^n (((x/y))^ +1)^n =y^n Σ_(k=0) ^n C_n ^k ((x/y))^k =Σ_(k=0) ^(n ) C_n ^k x^k y^n y^(−k) =Σ_(k=0) ^n C_n ^k x^k y^(n−k) ...the relation of binome is proved .$
$l e t p r o v e f i r s t t h a t {(x + 1)}^{n} = \sum_{k = 0}^{n} C_{n}^{k} x^{k} w i t h C_{n}^{k} = \frac{n!}{k! (n - k)!} b y r e c u r r e n c e$ $l e t p u t P_{n} (x) = {(x + 1)}^{n} - \sum_{k = 0}^{n} C_{n}^{k} x^{k} l e t p r o v e t h a t P_{n} (x) = 0$ $P_{0} (x) = 0 (t r u e) l e t s u p p o s e P_{n} (x) = 0 w e h a v e$ $P_{n + 1} (x) = {(x + 1)}^{n + 1} - \sum_{k = 0}^{n + 1} C_{n + 1}^{k} x^{k}$ $= {(x + 1)}^{n} (x + 1) - \sum_{k = 0}^{n + 1} C_{n + 1}^{k} x^{k} = (x + 1) {P_{n} (x) + \sum_{k = 0}^{n} C_{n}^{k} x^{k}} - \sum_{k = 0}^{n + 1} C_{n + 1}^{k} x^{k} b u t P_{n} (x) = 0 \Rightarrow$ $P_{n + 1} (x) = (x + 1) \sum_{k = 0}^{n} C_{n}^{k} x^{k} - 1 - \sum_{k = 1}^{n + 1} {C_{n}^{k} + C_{n}^{k - 1}} x^{k}$ $= x \sum_{k = 0}^{n} C_{n}^{k} x^{k} + \sum_{k = 0}^{n} C_{n}^{k} x^{k} - 1 - \sum_{k = 1}^{n} C_{n}^{k} x^{k} - \sum_{k = 1}^{n + 1} C_{n}^{k - 1} x^{k}$ $= x \sum_{k = 0}^{n} C_{n}^{k} x^{k} - \sum_{p = 0}^{n} C_{n}^{p} x^{p + 1} (c h a n g e m e n t k - 1 = p)$ $= x (\sum_{k = 0}^{n} C_{n}^{k} x^{k} - \sum_{p = 0}^{n} C_{n}^{p} x^{p}) = 0 \Rightarrow P_{n + 1} (x) = 0 s o$ ${(x + 1)}^{n} = \sum_{k = 0}^{n} C_{n}^{k} x^{k} w e h a v e$ ${(x + y)}^{n} = x^{n} i f y = 0 l e t s u p p o s e y \neq 0 \Rightarrow$ ${(x + y)}^{n} = y^{n} {({(\frac{x}{y})}^{} + 1)}^{n} = y^{n} \sum_{k = 0}^{n} C_{n}^{k} {(\frac{x}{y})}^{k} = \sum_{k = 0}^{n} C_{n}^{k} x^{k} y^{n} y^{- k}$ $= \sum_{k = 0}^{n} C_{n}^{k} x^{k} y^{n - k} . . . t h e r e l a t i o n o f b i n o m e i s p r o v e d .$
Commented by Hassen_Timol last updated on 07/Apr/19

$T h a n k y o u S i r$
Commented by maxmathsup by imad last updated on 07/Apr/19

$y o u a r e w e l c o m e .$
	Answered by 121194 last updated on 06/Apr/19

	$this just the binomial theorem$ $n! = n (n - 1)!$ $0! = 1$ $Σ denotes sum$ ${(x + y)}^{2} = \underset{k = 0}{\sum^{2}} (\begin{matrix} 2 \\ k \end{matrix}) x^{2 - k} y^{k}$ $= (\begin{matrix} 2 \\ 0 \end{matrix}) x^{2} y^{0} + (\begin{matrix} 2 \\ 1 \end{matrix}) x^{1} y^{1} + (\begin{matrix} 2 \\ 2 \end{matrix}) x^{0} y^{2}$ $= \frac{2!}{0! 2!} x^{2} + \frac{2!}{1! 1!} x y + \frac{2!}{2! 0!} y^{2}$ $= x^{2} + 2 x y + y^{2}$ $- - - - - - - - - - - - - - - - - - - - - - - -$ ${(x + y)}^{2} = (x + y) (x + y)$ $= x^{2} + x y + y x + y^{2}$ $= x^{2} + 2 x y + y^{2}$ $- - - - - - - - - - - - - - - - - - - - - - - -$ $there a combinatory argument to arive$ $it, like choosing k x factors from all the$ $(x + y) . . . . (x + y)$ $where k vary from 0 (no x factor get) to$ $n (all are x factor)$ $wich gives (\begin{matrix} n \\ k \end{matrix}) ways to do x^{k} y^{n - k}$ $btw this {don}^{'} t work with matrix because they$ $not commute in general .$
	Commented by Hassen_Timol last updated on 07/Apr/19

	$T h a n k y o u S i r$
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