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Question Number 3411 by Filup last updated on 13/Dec/15
Can someone show me how:  (i)    (x+a)^v =Σ_(k=0) ^v  ((v),(k) ) x^(v−k) a^k   (ii)  Does:       Σ_(k=0) ^v  ((v),(k) ) x^(v−k) a^k =Σ_(k=0) ^v  ((v),(k) ) x^k a^(v−k)
$$\mathrm{Can}\:\mathrm{someone}\:\mathrm{show}\:\mathrm{me}\:\mathrm{how}: \\ $$$$\left({i}\right)\:\:\:\:\left({x}+{a}\right)^{{v}} =\underset{{k}=\mathrm{0}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{k}}\end{pmatrix}\:{x}^{{v}−{k}} {a}^{{k}} \\ $$$$\left({ii}\right)\:\:\mathrm{Does}: \\ $$$$\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{k}}\end{pmatrix}\:{x}^{{v}−{k}} {a}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{k}}\end{pmatrix}\:{x}^{{k}} {a}^{{v}−{k}} \\ $$
Commented by RasheedSindhi last updated on 13/Dec/15
 ((v),(k) )   is defined in case k≤v only  (i) Can be proved by mathematical  induction if k≤v .  (ii) as x+a=a+x  ∴ Σ_(k=0) ^(v)  ((v),(k) ) x^(v−k) a^k =Σ_(k=0) ^v  ((v),(k) ) x^k a^(v−k)   (x and a are exchanged)
$$\begin{pmatrix}{{v}}\\{{k}}\end{pmatrix}\:\:\:{is}\:{defined}\:{in}\:{case}\:{k}\leqslant{v}\:{only} \\ $$$$\left({i}\right)\:\mathcal{C}{an}\:{be}\:{proved}\:{by}\:{mathematical} \\ $$$${induction}\:{if}\:{k}\leqslant{v}\:. \\ $$$$\left({ii}\right)\:{as}\:{x}+{a}={a}+{x} \\ $$$$\therefore\:\overset{{v}} {\underset{{k}=\mathrm{0}} {\sum}}\begin{pmatrix}{{v}}\\{{k}}\end{pmatrix}\:{x}^{{v}−{k}} {a}^{{k}} =\underset{{k}=\mathrm{0}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{k}}\end{pmatrix}\:{x}^{{k}} {a}^{{v}−{k}} \\ $$$$\left({x}\:{and}\:{a}\:{are}\:{exchanged}\right) \\ $$$$ \\ $$
Commented by Filup last updated on 13/Dec/15
Thanks. I thought as much for (ii).  I don′t know anything about   mathematical induction.
$$\mathrm{Thanks}.\:\mathrm{I}\:\mathrm{thought}\:\mathrm{as}\:\mathrm{much}\:\mathrm{for}\:\left({ii}\right). \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{anything}\:\mathrm{about}\: \\ $$$$\mathrm{mathematical}\:\mathrm{induction}. \\ $$
Commented by 123456 last updated on 13/Dec/15
in some places its defined to be 0   ((a),(b) )=((a!)/(b!(a−b)!))   ((a),(b) )=0,a<b,a∈N,b∈N
$$\mathrm{in}\:\mathrm{some}\:\mathrm{places}\:\mathrm{its}\:\mathrm{defined}\:\mathrm{to}\:\mathrm{be}\:\mathrm{0} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}=\frac{{a}!}{{b}!\left({a}−{b}\right)!} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}=\mathrm{0},{a}<{b},{a}\in\mathbb{N},{b}\in\mathbb{N} \\ $$
Commented by RasheedSindhi last updated on 13/Dec/15
This is knowledge for me!  for a=b   ((a),(b) ) =1 or 0?
$$\mathcal{T}{his}\:{is}\:{knowledge}\:{for}\:{me}! \\ $$$${for}\:{a}={b}\:\:\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}\:=\mathrm{1}\:{or}\:\mathrm{0}? \\ $$
Commented by 123456 last updated on 13/Dec/15
1, i aciddentaly write it wrong
$$\mathrm{1},\:\mathrm{i}\:\mathrm{aciddentaly}\:\mathrm{write}\:\mathrm{it}\:\mathrm{wrong} \\ $$
Commented by Filup last updated on 13/Dec/15
 ((a),(a) ) = ((a!)/((a−a)! a!))  =(1/1)
$$\begin{pmatrix}{{a}}\\{{a}}\end{pmatrix}\:=\:\frac{{a}!}{\left({a}−{a}\right)!\:{a}!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}} \\ $$
Commented by Yozzi last updated on 13/Dec/15
There is also a combinatorical  approach to proving the Binomial  theorem.
$${There}\:{is}\:{also}\:{a}\:{combinatorical} \\ $$$${approach}\:{to}\:{proving}\:{the}\:{Binomial} \\ $$$${theorem}. \\ $$
Commented by 123456 last updated on 13/Dec/15
for curiosity there some reaction to take   ((a),(b) )=0    a∈N∧b∈Z∧(b<0∨b>a)  we have   ((a),(b) )=((a!)/(b!(a−b)!))=((Γ(a+1))/(Γ(b+1)Γ(a−b+1)))  x!=Γ(x+1)  the poles of Γ(x) are {...,−2,−1,0}  so the poles of x! are {...,−3,−2,−1}  so  if b<0 we get a pole at b!  if a<b⇒a−b<0 we get a pole at (a−b)!  so its act like  ((a!)/(x!(a−x)!))=(k_1 /(±∞k_2 ))=0  however this is if we take the result of  limit as (a,b)  lim_((x,y)→(a,b))  ((x),(y) )  you can also compute it for non intergers by gamma function   ((1),((1/2)) )=((1!)/([(1/2)!]^2 ))=(4/π)
$$\mathrm{for}\:\mathrm{curiosity}\:\mathrm{there}\:\mathrm{some}\:\mathrm{reaction}\:\mathrm{to}\:\mathrm{take} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}=\mathrm{0}\:\:\:\:{a}\in\mathbb{N}\wedge{b}\in\mathbb{Z}\wedge\left({b}<\mathrm{0}\vee{b}>{a}\right) \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}=\frac{{a}!}{{b}!\left({a}−{b}\right)!}=\frac{\Gamma\left({a}+\mathrm{1}\right)}{\Gamma\left({b}+\mathrm{1}\right)\Gamma\left({a}−{b}+\mathrm{1}\right)} \\ $$$${x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\Gamma\left({x}\right)\:\mathrm{are}\:\left\{…,−\mathrm{2},−\mathrm{1},\mathrm{0}\right\} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:{x}!\:\mathrm{are}\:\left\{…,−\mathrm{3},−\mathrm{2},−\mathrm{1}\right\} \\ $$$$\mathrm{so} \\ $$$$\mathrm{if}\:{b}<\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{pole}\:\mathrm{at}\:{b}! \\ $$$$\mathrm{if}\:{a}<{b}\Rightarrow{a}−{b}<\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{pole}\:\mathrm{at}\:\left({a}−{b}\right)! \\ $$$$\mathrm{so}\:\mathrm{its}\:\mathrm{act}\:\mathrm{like} \\ $$$$\frac{{a}!}{{x}!\left({a}−{x}\right)!}=\frac{{k}_{\mathrm{1}} }{\pm\infty{k}_{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{however}\:\mathrm{this}\:\mathrm{is}\:\mathrm{if}\:\mathrm{we}\:\mathrm{take}\:\mathrm{the}\:\mathrm{result}\:\mathrm{of} \\ $$$$\mathrm{limit}\:\mathrm{as}\:\left({a},{b}\right) \\ $$$$\underset{\left({x},{y}\right)\rightarrow\left({a},{b}\right)} {\mathrm{lim}}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{also}\:\mathrm{compute}\:\mathrm{it}\:\mathrm{for}\:\mathrm{non}\:\mathrm{intergers}\:\mathrm{by}\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}/\mathrm{2}}\end{pmatrix}=\frac{\mathrm{1}!}{\left[\left(\mathrm{1}/\mathrm{2}\right)!\right]^{\mathrm{2}} }=\frac{\mathrm{4}}{\pi} \\ $$
Answered by 123456 last updated on 13/Dec/15
lets shown (x+a)^n =Σ_(k=0) ^n  ((n),(k) )x^(n−k) a^k    n∈N  base case:  n=0  (x+a)^0 =1  n=1  (x+a)^1 =x+a  inductive steep:  assuming its truth to n, let shown to  n+1  (x+a)^(n+1) =(x+a)(x+a)^n   =(x+a)Σ_(k=0) ^n  ((n),(k) )x^(n−k) a^k   =Σ_(k=0) ^n  ((n),(k) )x^(n+1−k) a^k +Σ_(k=0) ^n  ((n),(k) )x^(n−k) a^(k+1)   =Σ_(k=0) ^n  ((n),(k) )x^(n+1−k) a^k +Σ_(k=1) ^(n+1)  ((n),((k−1)) )x^(n+1−k) a^k   =x^(n+1) +Σ_(k=1) ^n  ((n),(k) )x^(n+1−k) a^k +Σ_(k=1) ^n  ((n),((k−1)) )x^(n+1−k) a^k +a^(n+1)   =x^(n+1) +Σ_(k=1) ^n ( ((n),(k) )+ ((n),((k−1)) ) )x^(n+1−k) a^k +a^(n+1)   =x^(n+1) +Σ_(k=1) ^n  (((n+1)),(k) )x^(n+1−k) a^k +a^(n+1)   =Σ_(k=0) ^(n+1)  (((n+1)),(k) )x^(n+1−k) a^k     □   ((n),(k) )+ ((n),((k−1)) )= (((n+1)),(k) )
$$\mathrm{lets}\:\mathrm{shown}\:\left({x}+{a}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{n}−{k}} {a}^{{k}} \:\:\:{n}\in\mathbb{N} \\ $$$$\mathrm{base}\:\mathrm{case}: \\ $$$${n}=\mathrm{0} \\ $$$$\left({x}+{a}\right)^{\mathrm{0}} =\mathrm{1} \\ $$$${n}=\mathrm{1} \\ $$$$\left({x}+{a}\right)^{\mathrm{1}} ={x}+{a} \\ $$$$\mathrm{inductive}\:\mathrm{steep}: \\ $$$$\mathrm{assuming}\:\mathrm{its}\:\mathrm{truth}\:\mathrm{to}\:{n},\:\mathrm{let}\:\mathrm{shown}\:\mathrm{to} \\ $$$${n}+\mathrm{1} \\ $$$$\left({x}+{a}\right)^{{n}+\mathrm{1}} =\left({x}+{a}\right)\left({x}+{a}\right)^{{n}} \\ $$$$=\left({x}+{a}\right)\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{n}−{k}} {a}^{{k}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} +\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{n}−{k}} {a}^{{k}+\mathrm{1}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} +\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}{x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} \\ $$$$={x}^{{n}+\mathrm{1}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}{x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}{x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} +{a}^{{n}+\mathrm{1}} \\ $$$$={x}^{{n}+\mathrm{1}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:\right){x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} +{a}^{{n}+\mathrm{1}} \\ $$$$={x}^{{n}+\mathrm{1}} +\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}{x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} +{a}^{{n}+\mathrm{1}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\sum}}\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}{x}^{{n}+\mathrm{1}−{k}} {a}^{{k}} \:\:\:\:\Box \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix} \\ $$
Commented by Yozzi last updated on 13/Dec/15
Proof of  ((n),(k) )+ ((n),((k−1)) )= (((n+1)),(k) ).  Suppose we a set of n distinct objects  and we choose one object as a special   one. Then, the number of sets of  k disinct objects is the sum of the   number of sets with the special object and   the number of sets without the special  object.   Given that we already took the special  object, we need k−1 objects (to add to  1 to give k objects) from n−1 objects  (we already took 1 object). This occurs  in  (((n−1)),((k−1)) ) ways where  ((p),(q) ) is the choose  function  ((p),(q) )=((p!)/((p−q)!q!)).  The number of sets without the special  object occurs in  (((n−1)),(k) ) ways since  we have only n−1 objects to choose k  objects from if the special object is  not to be part of the set.  So we have    ((n),(k) )= (((n−1)),((k−1)) )+ (((n−1)),(k) )  ⇒ for a group of n+1 distinct objects   (((n+1)),(k) )= ((n),((k−1)) )+ ((n),(k) ).                   □  We can likewise show that   ((n),(k) )= (((n−1)),(k) )+2 (((n−1)),((k−1)) )+ (((n−1)),((k−2)) ) , n,k∈Z^≥   n≥k.
$${Proof}\:{of}\:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}. \\ $$$${Suppose}\:{we}\:{a}\:{set}\:{of}\:{n}\:{distinct}\:{objects} \\ $$$${and}\:{we}\:{choose}\:{one}\:{object}\:{as}\:{a}\:{special}\: \\ $$$${one}.\:{Then},\:{the}\:{number}\:{of}\:{sets}\:{of} \\ $$$${k}\:{disinct}\:{objects}\:{is}\:{the}\:{sum}\:{of}\:{the}\: \\ $$$${number}\:{of}\:{sets}\:{with}\:{the}\:{special}\:{object}\:{and}\: \\ $$$${the}\:{number}\:{of}\:{sets}\:{without}\:{the}\:{special} \\ $$$${object}.\: \\ $$$${Given}\:{that}\:{we}\:{already}\:{took}\:{the}\:{special} \\ $$$${object},\:{we}\:{need}\:{k}−\mathrm{1}\:{objects}\:\left({to}\:{add}\:{to}\right. \\ $$$$\left.\mathrm{1}\:{to}\:{give}\:{k}\:{objects}\right)\:{from}\:{n}−\mathrm{1}\:{objects} \\ $$$$\left({we}\:{already}\:{took}\:\mathrm{1}\:{object}\right).\:{This}\:{occurs} \\ $$$${in}\:\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix}\:{ways}\:{where}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:{is}\:{the}\:{choose} \\ $$$${function}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}=\frac{{p}!}{\left({p}−{q}\right)!{q}!}. \\ $$$${The}\:{number}\:{of}\:{sets}\:{without}\:{the}\:{special} \\ $$$${object}\:{occurs}\:{in}\:\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}}\end{pmatrix}\:{ways}\:{since} \\ $$$${we}\:{have}\:{only}\:{n}−\mathrm{1}\:{objects}\:{to}\:{choose}\:{k} \\ $$$${objects}\:{from}\:{if}\:{the}\:{special}\:{object}\:{is} \\ $$$${not}\:{to}\:{be}\:{part}\:{of}\:{the}\:{set}. \\ $$$${So}\:{we}\:{have}\: \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}=\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}}\end{pmatrix} \\ $$$$\Rightarrow\:{for}\:{a}\:{group}\:{of}\:{n}+\mathrm{1}\:{distinct}\:{objects} \\ $$$$\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}=\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\ $$$${We}\:{can}\:{likewise}\:{show}\:{that} \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}=\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}}\end{pmatrix}+\mathrm{2}\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}−\mathrm{1}}\end{pmatrix}+\begin{pmatrix}{{n}−\mathrm{1}}\\{{k}−\mathrm{2}}\end{pmatrix}\:,\:{n},{k}\in\mathbb{Z}^{\geqslant} \\ $$$${n}\geqslant{k}. \\ $$
Answered by Yozzi last updated on 13/Dec/15
(x+a)^v =(a{1+(x/a)})^v   (x+a)^v =a^v {1+(x/a)}^v    Here, I assume that v∈Z^≥ . a≠0  (Z^≥ ={non−negative integers})  The choose function is given as   ((n),(r) )  as the number of ways of choosing  r objects from n distinct objects.    Of course we have  {1+(x/a)}^v =(1+(x/a))(1+(x/a))(1+(x/a))...(1+(x/a))                        (v times)  Here you have v brackets with (x/a) in  each of them so that powers of (x/a)   can arise in a number of ways as a  consequence of picking brackets to   multiply with each other to give the  sought after power of (x/a).  ((x/a))^v : To obtain this term, you have                v brackets to choose (x/a) from and                 this is done in  ((v),(v) ) ways. So,                 ((x/a))^v occurs  ((v),(v) ) times and thus                  the term in ((x/a))^v  is  ((v),(v) )((x/a))^v .  ((x/a))^(v−1) : Here it is the case that you′d                     like to obtain (x/a) multiplied                     by itself v−1 times.                     You have v brackets to choose                      from, so that this occurs                      in  ((v),((v−1)) ) ways. Hence, the                        term in ((x/a))^(v−1)   is  ((v),((v−1)) ) ((x/a))^(v−1) .  ((x/a))^(v−2) : This term arises in  ((v),((v−2)) ) ways                     from among the brackets, so                     this term is  ((v),((v−2)) )((x/a))^(v−2) .  Continuing on, terms in ((x/a))^2    arise in   ((v),(2) ) ways, terms in (x/a) arise  in  ((v),(1) ) ways and terms in ((x/a))^0  arise  in  ((v),(0) )=1 ways from among the brackets.  We hence obtain in all,  (x+a)^v =a^v [ ((v),(v) )((x/a))^v + ((v),((v−1)) )((x/a))^(v−1)                       + ((v),((v−2)) )((x/a))^(v−2) +...+ ((v),(2) )((x/a))^2 + ((v),(1) )((x/a))                       +1]  (x+a)^v =a^v (Σ_(r=0) ^v  ((v),(r) )x^r a^(−r) )  (x+a)^v =Σ_(r=0) ^v  ((v),(r) )x^r a^(v−r)                           □  If one approached the proof starting  with ′The number of ways of obtaining  the term ((a/x))^0  from {1+(a/x)}^v  (x≠0)  (after writting {x+a}^v =x^v (1+(a/x))^v )  is   ((v),(0) )...′ you get the equivalent   expression  (x+a)^v =Σ_(r=0) ^v  ((v),(r) )x^(v−r) a^r .
$$\left({x}+{a}\right)^{{v}} =\left({a}\left\{\mathrm{1}+\frac{{x}}{{a}}\right\}\right)^{{v}} \\ $$$$\left({x}+{a}\right)^{{v}} ={a}^{{v}} \left\{\mathrm{1}+\frac{{x}}{{a}}\right\}^{{v}} \: \\ $$$${Here},\:{I}\:{assume}\:{that}\:{v}\in\mathbb{Z}^{\geqslant} .\:{a}\neq\mathrm{0} \\ $$$$\left(\mathbb{Z}^{\geqslant} =\left\{{non}−{negative}\:{integers}\right\}\right) \\ $$$${The}\:{choose}\:{function}\:{is}\:{given}\:{as}\:\:\begin{pmatrix}{{n}}\\{{r}}\end{pmatrix} \\ $$$${as}\:{the}\:{number}\:{of}\:{ways}\:{of}\:{choosing} \\ $$$${r}\:{objects}\:{from}\:{n}\:{distinct}\:{objects}. \\ $$$$ \\ $$$${Of}\:{course}\:{we}\:{have} \\ $$$$\left\{\mathrm{1}+\frac{{x}}{{a}}\right\}^{{v}} =\left(\mathrm{1}+\frac{{x}}{{a}}\right)\left(\mathrm{1}+\frac{{x}}{{a}}\right)\left(\mathrm{1}+\frac{{x}}{{a}}\right)…\left(\mathrm{1}+\frac{{x}}{{a}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({v}\:{times}\right) \\ $$$${Here}\:{you}\:{have}\:{v}\:{brackets}\:{with}\:\frac{{x}}{{a}}\:{in} \\ $$$${each}\:{of}\:{them}\:{so}\:{that}\:{powers}\:{of}\:\frac{{x}}{{a}}\: \\ $$$${can}\:{arise}\:{in}\:{a}\:{number}\:{of}\:{ways}\:{as}\:{a} \\ $$$${consequence}\:{of}\:{picking}\:{brackets}\:{to}\: \\ $$$${multiply}\:{with}\:{each}\:{other}\:{to}\:{give}\:{the} \\ $$$${sought}\:{after}\:{power}\:{of}\:\frac{{x}}{{a}}. \\ $$$$\left(\frac{{x}}{{a}}\right)^{{v}} :\:{To}\:{obtain}\:{this}\:{term},\:{you}\:{have} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{v}\:{brackets}\:{to}\:{choose}\:\frac{{x}}{{a}}\:{from}\:{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{this}\:{is}\:{done}\:{in}\:\begin{pmatrix}{{v}}\\{{v}}\end{pmatrix}\:{ways}.\:{So}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{{x}}{{a}}\right)^{{v}} {occurs}\:\begin{pmatrix}{{v}}\\{{v}}\end{pmatrix}\:{times}\:{and}\:{thus} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{the}\:{term}\:{in}\:\left(\frac{{x}}{{a}}\right)^{{v}} \:{is}\:\begin{pmatrix}{{v}}\\{{v}}\end{pmatrix}\left(\frac{{x}}{{a}}\right)^{{v}} . \\ $$$$\left(\frac{{x}}{{a}}\right)^{{v}−\mathrm{1}} :\:{Here}\:{it}\:{is}\:{the}\:{case}\:{that}\:{you}'{d} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{like}\:{to}\:{obtain}\:\frac{{x}}{{a}}\:{multiplied} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{by}\:{itself}\:{v}−\mathrm{1}\:{times}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{You}\:{have}\:{v}\:{brackets}\:{to}\:{choose} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{from},\:{so}\:{that}\:{this}\:{occurs} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{in}\:\begin{pmatrix}{{v}}\\{{v}−\mathrm{1}}\end{pmatrix}\:{ways}.\:{Hence},\:{the}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{term}\:{in}\:\left(\frac{{x}}{{a}}\right)^{{v}−\mathrm{1}} \:\:{is}\:\begin{pmatrix}{{v}}\\{{v}−\mathrm{1}}\end{pmatrix}\:\left(\frac{{x}}{{a}}\right)^{{v}−\mathrm{1}} . \\ $$$$\left(\frac{{x}}{{a}}\right)^{{v}−\mathrm{2}} :\:{This}\:{term}\:{arises}\:{in}\:\begin{pmatrix}{{v}}\\{{v}−\mathrm{2}}\end{pmatrix}\:{ways} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{from}\:{among}\:{the}\:{brackets},\:{so} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{this}\:{term}\:{is}\:\begin{pmatrix}{{v}}\\{{v}−\mathrm{2}}\end{pmatrix}\left(\frac{{x}}{{a}}\right)^{{v}−\mathrm{2}} . \\ $$$${Continuing}\:{on},\:{terms}\:{in}\:\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} \: \\ $$$${arise}\:{in}\:\:\begin{pmatrix}{{v}}\\{\mathrm{2}}\end{pmatrix}\:{ways},\:{terms}\:{in}\:\frac{{x}}{{a}}\:{arise} \\ $$$${in}\:\begin{pmatrix}{{v}}\\{\mathrm{1}}\end{pmatrix}\:{ways}\:{and}\:{terms}\:{in}\:\left(\frac{{x}}{{a}}\right)^{\mathrm{0}} \:{arise} \\ $$$${in}\:\begin{pmatrix}{{v}}\\{\mathrm{0}}\end{pmatrix}=\mathrm{1}\:{ways}\:{from}\:{among}\:{the}\:{brackets}. \\ $$$${We}\:{hence}\:{obtain}\:{in}\:{all}, \\ $$$$\left({x}+{a}\right)^{{v}} ={a}^{{v}} \left[\begin{pmatrix}{{v}}\\{{v}}\end{pmatrix}\left(\frac{{x}}{{a}}\right)^{{v}} +\begin{pmatrix}{{v}}\\{{v}−\mathrm{1}}\end{pmatrix}\left(\frac{{x}}{{a}}\right)^{{v}−\mathrm{1}} \right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\begin{pmatrix}{{v}}\\{{v}−\mathrm{2}}\end{pmatrix}\left(\frac{{x}}{{a}}\right)^{{v}−\mathrm{2}} +…+\begin{pmatrix}{{v}}\\{\mathrm{2}}\end{pmatrix}\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} +\begin{pmatrix}{{v}}\\{\mathrm{1}}\end{pmatrix}\left(\frac{{x}}{{a}}\right) \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{1}\right] \\ $$$$\left({x}+{a}\right)^{{v}} ={a}^{{v}} \left(\underset{{r}=\mathrm{0}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{r}}\end{pmatrix}{x}^{{r}} {a}^{−{r}} \right) \\ $$$$\left({x}+{a}\right)^{{v}} =\underset{{r}=\mathrm{0}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{r}}\end{pmatrix}{x}^{{r}} {a}^{{v}−{r}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\ $$$${If}\:{one}\:{approached}\:{the}\:{proof}\:{starting} \\ $$$${with}\:'{The}\:{number}\:{of}\:{ways}\:{of}\:{obtaining} \\ $$$${the}\:{term}\:\left(\frac{{a}}{{x}}\right)^{\mathrm{0}} \:{from}\:\left\{\mathrm{1}+\frac{{a}}{{x}}\right\}^{{v}} \:\left({x}\neq\mathrm{0}\right) \\ $$$$\left({after}\:{writting}\:\left\{{x}+{a}\right\}^{{v}} ={x}^{{v}} \left(\mathrm{1}+\frac{{a}}{{x}}\right)^{{v}} \right) \\ $$$${is}\:\:\begin{pmatrix}{{v}}\\{\mathrm{0}}\end{pmatrix}…'\:{you}\:{get}\:{the}\:{equivalent}\: \\ $$$${expression} \\ $$$$\left({x}+{a}\right)^{{v}} =\underset{{r}=\mathrm{0}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{r}}\end{pmatrix}{x}^{{v}−{r}} {a}^{{r}} . \\ $$
Answered by prakash jain last updated on 13/Dec/15
f(x)=(x+a)^n   f ′(x)=n(x+a)^(n−1)   f^((2)) (x)=n(n−1)(x+a)^(n−2)   f^((n)) (x)=n!  f^((k)) (x)=0, k>n  Taylor series  f(x)=a^n +(n/(1!))x+((n(n−1))/(2!))x^2 +...+((n!)/(n!))x^n +0x^(n+1) +..  (x+a)^n =Σ_(i=0) ^n ^n C_i x^i a^(n−i)
$${f}\left({x}\right)=\left({x}+{a}\right)^{{n}} \\ $$$${f}\:'\left({x}\right)={n}\left({x}+{a}\right)^{{n}−\mathrm{1}} \\ $$$${f}^{\left(\mathrm{2}\right)} \left({x}\right)={n}\left({n}−\mathrm{1}\right)\left({x}+{a}\right)^{{n}−\mathrm{2}} \\ $$$${f}^{\left({n}\right)} \left({x}\right)={n}! \\ $$$${f}^{\left({k}\right)} \left({x}\right)=\mathrm{0},\:{k}>{n} \\ $$$$\mathrm{Taylor}\:\mathrm{series} \\ $$$${f}\left({x}\right)={a}^{{n}} +\frac{{n}}{\mathrm{1}!}{x}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +…+\frac{{n}!}{{n}!}{x}^{{n}} +\mathrm{0}{x}^{{n}+\mathrm{1}} +.. \\ $$$$\left({x}+{a}\right)^{{n}} =\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{i}} {x}^{{i}} {a}^{{n}−{i}} \\ $$

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