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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
Excuse me, what does it mean? Can you give some examples for n?
$${Excuse}\:{me},\:{what}\:{does}\:{it}\:{mean}? \\ $$$${Can}\:{you}\:{give}\:{some}\:{examples}\:{for}\:\mathrm{n}? \\ $$
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 .
$${let}\:{prove}\:\:{first}\:{that}\:\left({x}+\mathrm{1}\right)^{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\:\:\:\:{with}\:{C}_{{n}} ^{{k}} \:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:\:{by}\:{recurrence} \\ $$$${let}\:{put}\:{P}_{{n}} \left({x}\right)\:=\left({x}+\mathrm{1}\right)^{{n}} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\:\:\:{let}\:{prove}\:{that}\:{P}_{{n}} \left({x}\right)=\mathrm{0}\:\: \\ $$$${P}_{\mathrm{0}} \left({x}\right)\:=\mathrm{0}\:\:\left({true}\right)\:\:\:{let}\:{suppose}\:{P}_{{n}} \left({x}\right)=\mathrm{0}\:\:{we}\:{have}\: \\ $$$${P}_{{n}+\mathrm{1}} \left({x}\right)\:=\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} \:−\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{C}_{{n}+\mathrm{1}} ^{{k}} \:{x}^{{k}} \: \\ $$$$=\left({x}+\mathrm{1}\right)^{{n}} \left({x}+\mathrm{1}\right)−\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{C}_{{n}+\mathrm{1}} ^{{k}} \:{x}^{{k}} \:\:=\left({x}+\mathrm{1}\right)\left\{{P}_{{n}} \left({x}\right)+\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \right\}−\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{C}_{{n}+\mathrm{1}} ^{{k}} \:{x}^{{k}} \:{but}\:{P}_{{n}} \left({x}\right)=\mathrm{0}\:\Rightarrow \\ $$$${P}_{{n}+\mathrm{1}} \left({x}\right)=\left({x}+\mathrm{1}\right)\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:−\mathrm{1}−\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:\left\{{C}_{{n}} ^{{k}} \:+{C}_{{n}} ^{{k}−\mathrm{1}} \right\}{x}^{{k}} \\ $$$$={x}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:+\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\:−\mathrm{1}\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:−\sum_{{k}=\mathrm{1}} ^{{n}+\mathrm{1}} \:{C}_{{n}} ^{{k}−\mathrm{1}} \:{x}^{{k}} \\ $$$$={x}\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:−\sum_{{p}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{p}} \:{x}^{{p}+\mathrm{1}} \left(\:{changement}\:{k}−\mathrm{1}={p}\right) \\ $$$$={x}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:−\sum_{{p}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{p}} \:{x}^{{p}} \right)\:=\mathrm{0}\:\Rightarrow{P}_{{n}+\mathrm{1}} \left({x}\right)=\mathrm{0}\:\:\:{so} \\ $$$$\left({x}+\mathrm{1}\right)^{{n}\:} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\:\:\:\:\:\:\:\:{we}\:{have} \\ $$$$\left({x}+{y}\right)^{{n}} \:={x}^{{n}} \:{if}\:{y}=\mathrm{0}\:{let}\:{suppose}\:{y}\neq\mathrm{0}\:\Rightarrow \\ $$$$\left({x}+{y}\right)^{{n}} \:={y}^{{n}} \left(\left(\frac{{x}}{{y}}\right)^{} \:+\mathrm{1}\right)^{{n}} \:\:={y}^{{n}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(\frac{{x}}{{y}}\right)^{{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}\:\:\:} \:{C}_{{n}} ^{{k}} \:\:{x}^{{k}} \:{y}^{{n}} \:{y}^{−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:{y}^{{n}−{k}} \:…{the}\:{relation}\:{of}\:{binome}\:{is}\:{proved}\:. \\ $$$$ \\ $$
Commented by Hassen_Timol last updated on 07/Apr/19
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$
Commented by maxmathsup by imad last updated on 07/Apr/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$
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 =Σ_(k=0) ^2 ((2),(k) )x^(2−k) y^k = ((2),(0) )x^2 y^0 + ((2),(1) )x^1 y^1 + ((2),(2) )x^0 y^2 =((2!)/(0!2!))x^2 +((2!)/(1!1!))xy+((2!)/(2!0!))y^2 =x^2 +2xy+y^2 −−−−−−−−−−−−−−−−−−−−−−−− (x+y)^2 =(x+y)(x+y) =x^2 +xy+yx+y^2 =x^2 +2xy+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 ((n),(k) ) ways to do x^k y^(n−k) btw this don′t work with matrix because they not commute in general.
$$\mathrm{this}\:\mathrm{just}\:\mathrm{the}\:\mathrm{binomial}\:\mathrm{theorem} \\ $$$${n}!={n}\left({n}−\mathrm{1}\right)! \\ $$$$\mathrm{0}!=\mathrm{1} \\ $$$$\Sigma\:\mathrm{denotes}\:\mathrm{sum} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}} {\sum}}\begin{pmatrix}{\mathrm{2}}\\{{k}}\end{pmatrix}{x}^{\mathrm{2}−{k}} {y}^{{k}} \\ $$$$=\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix}{x}^{\mathrm{2}} {y}^{\mathrm{0}} +\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}{x}^{\mathrm{1}} {y}^{\mathrm{1}} +\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}{x}^{\mathrm{0}} {y}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}!}{\mathrm{0}!\mathrm{2}!}{x}^{\mathrm{2}} +\frac{\mathrm{2}!}{\mathrm{1}!\mathrm{1}!}{xy}+\frac{\mathrm{2}!}{\mathrm{2}!\mathrm{0}!}{y}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} =\left({x}+{y}\right)\left({x}+{y}\right) \\ $$$$={x}^{\mathrm{2}} +{xy}+{yx}+{y}^{\mathrm{2}} \\ $$$$={x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{there}\:\mathrm{a}\:\mathrm{combinatory}\:\mathrm{argument}\:\mathrm{to}\:\mathrm{arive} \\ $$$$\mathrm{it},\:\mathrm{like}\:\mathrm{choosing}\:{k}\:{x}\:\mathrm{factors}\:\mathrm{from}\:\mathrm{all}\:\mathrm{the} \\ $$$$\left({x}+{y}\right)….\left({x}+{y}\right) \\ $$$$\mathrm{where}\:{k}\:\mathrm{vary}\:\mathrm{from}\:\mathrm{0}\:\left(\mathrm{no}\:{x}\:\mathrm{factor}\:\mathrm{get}\right)\:\mathrm{to} \\ $$$${n}\:\left(\mathrm{all}\:\mathrm{are}\:{x}\:\mathrm{factor}\right) \\ $$$$\mathrm{wich}\:\mathrm{gives}\:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:\mathrm{ways}\:\mathrm{to}\:\mathrm{do}\:{x}^{{k}} {y}^{{n}−{k}} \\ $$$$\mathrm{btw}\:\mathrm{this}\:\mathrm{don}'\mathrm{t}\:\mathrm{work}\:\mathrm{with}\:\mathrm{matrix}\:\mathrm{because}\:\mathrm{they} \\ $$$$\mathrm{not}\:\mathrm{commute}\:\mathrm{in}\:\mathrm{general}. \\ $$
Commented by Hassen_Timol last updated on 07/Apr/19
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$

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