# Can-someone-show-me-how-i-x-a-v-k-0-v-v-k-x-v-k-a-k-ii-Does-k-0-v-v-k-x-v-k-a-k-k-0-v-v-k-x-k-a-v-k-

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