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


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Question Number 3411 by Filup last updated on 13/Dec/15

$Can someone show me how :$ $(i) {(x + a)}^{v} = \underset{k = 0}{\sum^{v}} (\begin{matrix} v \\ k \end{matrix}) x^{v - k} a^{k}$ $(i i) Does :$ $\underset{k = 0}{\sum^{v}} (\begin{matrix} v \\ k \end{matrix}) x^{v - k} a^{k} = \underset{k = 0}{\sum^{v}} (\begin{matrix} v \\ k \end{matrix}) x^{k} a^{v - k}$
Commented by RasheedSindhi last updated on 13/Dec/15

$(\begin{matrix} v \\ k \end{matrix}) i s d e f i n e d i n c a s e k ⩽ v o n l y$ $(i) C a n b e p r o v e d b y m a t h e m a t i c a l$ $i n d u c t i o n i f k ⩽ v .$ $(i i) a s x + a = a + x$ $∴ \overset{v}{\sum_{k = 0}} (\begin{matrix} v \\ k \end{matrix}) x^{v - k} a^{k} = \underset{k = 0}{\sum^{v}} (\begin{matrix} v \\ k \end{matrix}) x^{k} a^{v - k}$ $(x a n d a a r e e x c h a n g e d)$
Commented by Filup last updated on 13/Dec/15

$Thanks . I thought as much for (i i) .$ $I {don}^{'} t know anything about$ $mathematical induction .$
Commented by 123456 last updated on 13/Dec/15

$in some places its defined to be 0$ $(\begin{matrix} a \\ b \end{matrix}) = \frac{a!}{b! (a - b)!}$ $(\begin{matrix} a \\ b \end{matrix}) = 0, a < b, a \in N, b \in N$
Commented by RasheedSindhi last updated on 13/Dec/15

$T h i s i s k n o w l e d g e f o r m e!$ $f o r a = b (\begin{matrix} a \\ b \end{matrix}) = 1 o r 0 ?$
Commented by 123456 last updated on 13/Dec/15

$1, i aciddentaly write it wrong$
Commented by Filup last updated on 13/Dec/15

$(\begin{matrix} a \\ a \end{matrix}) = \frac{a!}{(a - a)! a!}$ $= \frac{1}{1}$
Commented by Yozzi last updated on 13/Dec/15

$T h e r e i s a l s o a c o m b i n a t o r i c a l$ $a p p r o a c h t o p r o v i n g t h e B i n o m i a l$ $t h e o r e m .$
Commented by 123456 last updated on 13/Dec/15
$for curiosity there some reaction to take ((a),(b) )=0 a∈N∧b∈Z∧(b<0∨b>a) we have ((a),(b) )=((a!)/(b!(a−b)!))=((Γ(a+1))/(Γ(b+1)Γ(a−b+1))) x!=Γ(x+1) the poles of Γ(x) are {...,−2,−1,0} so the poles of x! are {...,−3,−2,−1} so if b<0 we get a pole at b! if a<b⇒a−b<0 we get a pole at (a−b)! so its act like ((a!)/(x!(a−x)!))=(k_1 /(±∞k_2 ))=0 however this is if we take the result of limit as (a,b) lim_((x,y)→(a,b)) ((x),(y) ) you can also compute it for non intergers by gamma function ((1),((1/2)) )=((1!)/([(1/2)!]^2 ))=(4/π)$
$for curiosity there some reaction to take$ $(\begin{matrix} a \\ b \end{matrix}) = 0 a \in N \land b \in Z \land (b < 0 \lor b > a)$ $we have$ $(\begin{matrix} a \\ b \end{matrix}) = \frac{a!}{b! (a - b)!} = \frac{Γ (a + 1)}{Γ (b + 1) Γ (a - b + 1)}$ $x! = Γ (x + 1)$ $the poles of Γ (x) are {. . ., - 2, - 1, 0}$ $so the poles of x! are {. . ., - 3, - 2, - 1}$ $so$ $if b < 0 we get a pole at b!$ $if a < b \Rightarrow a - b < 0 we get a pole at (a - b)!$ $so its act like$ $\frac{a!}{x! (a - x)!} = \frac{k_{1}}{\pm \infty k_{2}} = 0$ $however this is if we take the result of$ $limit as (a, b)$ $\lim_{(x, y) \to (a, b)} (\begin{matrix} x \\ y \end{matrix})$ $you can also compute it for non intergers by gamma function$ $(\begin{matrix} 1 \\ 1 / 2 \end{matrix}) = \frac{1!}{{[(1 / 2)!]}^{2}} = \frac{4}{π}$
	Answered by 123456 last updated on 13/Dec/15

	$lets shown {(x + a)}^{n} = \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} a^{k} n \in N$ $base case :$ $n = 0$ ${(x + a)}^{0} = 1$ $n = 1$ ${(x + a)}^{1} = x + a$ $inductive steep :$ $assuming its truth to n, let shown to$ $n + 1$ ${(x + a)}^{n + 1} = (x + a) {(x + a)}^{n}$ $= (x + a) \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} a^{k}$ $= \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) x^{n + 1 - k} a^{k} + \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) x^{n - k} a^{k + 1}$ $= \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) x^{n + 1 - k} a^{k} + \underset{k = 1}{\sum^{n + 1}} (\begin{matrix} n \\ k - 1 \end{matrix}) x^{n + 1 - k} a^{k}$ $= x^{n + 1} + \underset{k = 1}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) x^{n + 1 - k} a^{k} + \underset{k = 1}{\sum^{n}} (\begin{matrix} n \\ k - 1 \end{matrix}) x^{n + 1 - k} a^{k} + a^{n + 1}$ $= x^{n + 1} + \underset{k = 1}{\sum^{n}} ((\begin{matrix} n \\ k \end{matrix}) + (\begin{matrix} n \\ k - 1 \end{matrix})) x^{n + 1 - k} a^{k} + a^{n + 1}$ $= x^{n + 1} + \underset{k = 1}{\sum^{n}} (\begin{matrix} n + 1 \\ k \end{matrix}) x^{n + 1 - k} a^{k} + a^{n + 1}$ $= \underset{k = 0}{\sum^{n + 1}} (\begin{matrix} n + 1 \\ k \end{matrix}) x^{n + 1 - k} a^{k} ◻$ $(\begin{matrix} n \\ k \end{matrix}) + (\begin{matrix} n \\ k - 1 \end{matrix}) = (\begin{matrix} n + 1 \\ k \end{matrix})$
	Commented by Yozzi last updated on 13/Dec/15

	$P r o o f o f (\begin{matrix} n \\ k \end{matrix}) + (\begin{matrix} n \\ k - 1 \end{matrix}) = (\begin{matrix} n + 1 \\ k \end{matrix}) .$ $S u p p o s e w e a s e t o f n d i s t i n c t o b j e c t s$ $a n d w e c h o o s e o n e o b j e c t a s a s p e c i a l$ $o n e . T h e n, t h e n u m b e r o f s e t s o f$ $k d i s i n c t o b j e c t s i s t h e s u m o f t h e$ $n u m b e r o f s e t s w i t h t h e s p e c i a l o b j e c t a n d$ $t h e n u m b e r o f s e t s w i t h o u t t h e s p e c i a l$ $o b j e c t .$ $G i v e n t h a t w e a l r e a d y t o o k t h e s p e c i a l$ $o b j e c t, w e n e e d k - 1 o b j e c t s (t o a d d t o$ $1 t o g i v e k o b j e c t s) f r o m n - 1 o b j e c t s$ $(w e a l r e a d y t o o k 1 o b j e c t) . T h i s o c c u r s$ $i n (\begin{matrix} n - 1 \\ k - 1 \end{matrix}) w a y s w h e r e (\begin{matrix} p \\ q \end{matrix}) i s t h e c h o o s e$ $f u n c t i o n (\begin{matrix} p \\ q \end{matrix}) = \frac{p!}{(p - q)! q!} .$ $T h e n u m b e r o f s e t s w i t h o u t t h e s p e c i a l$ $o b j e c t o c c u r s i n (\begin{matrix} n - 1 \\ k \end{matrix}) w a y s s i n c e$ $w e h a v e o n l y n - 1 o b j e c t s t o c h o o s e k$ $o b j e c t s f r o m i f t h e s p e c i a l o b j e c t i s$ $n o t t o b e p a r t o f t h e s e t .$ $S o w e h a v e$ $(\begin{matrix} n \\ k \end{matrix}) = (\begin{matrix} n - 1 \\ k - 1 \end{matrix}) + (\begin{matrix} n - 1 \\ k \end{matrix})$ $\Rightarrow f o r a g r o u p o f n + 1 d i s t i n c t o b j e c t s$ $(\begin{matrix} n + 1 \\ k \end{matrix}) = (\begin{matrix} n \\ k - 1 \end{matrix}) + (\begin{matrix} n \\ k \end{matrix}) . ◻$ $W e c a n l i k e w i s e s h o w t h a t$ $(\begin{matrix} n \\ k \end{matrix}) = (\begin{matrix} n - 1 \\ k \end{matrix}) + 2 (\begin{matrix} n - 1 \\ k - 1 \end{matrix}) + (\begin{matrix} n - 1 \\ k - 2 \end{matrix}), n, k \in Z^{⩾}$ $n ⩾ k .$
	Answered by Yozzi last updated on 13/Dec/15
	$(x+a)^v =(a{1+(x/a)})^v (x+a)^v =a^v {1+(x/a)}^v Here, I assume that v∈Z^≥ . a≠0 (Z^≥ ={non−negative integers}) The choose function is given as ((n),(r) ) as the number of ways of choosing r objects from n distinct objects. Of course we have {1+(x/a)}^v =(1+(x/a))(1+(x/a))(1+(x/a))...(1+(x/a)) (v times) Here you have v brackets with (x/a) in each of them so that powers of (x/a) can arise in a number of ways as a consequence of picking brackets to multiply with each other to give the sought after power of (x/a). ((x/a))^v : To obtain this term, you have v brackets to choose (x/a) from and this is done in ((v),(v) ) ways. So, ((x/a))^v occurs ((v),(v) ) times and thus the term in ((x/a))^v is ((v),(v) )((x/a))^v . ((x/a))^(v−1) : Here it is the case that you′d like to obtain (x/a) multiplied by itself v−1 times. You have v brackets to choose from, so that this occurs in ((v),((v−1)) ) ways. Hence, the term in ((x/a))^(v−1) is ((v),((v−1)) ) ((x/a))^(v−1) . ((x/a))^(v−2) : This term arises in ((v),((v−2)) ) ways from among the brackets, so this term is ((v),((v−2)) )((x/a))^(v−2) . Continuing on, terms in ((x/a))^2 arise in ((v),(2) ) ways, terms in (x/a) arise in ((v),(1) ) ways and terms in ((x/a))^0 arise in ((v),(0) )=1 ways from among the brackets. We hence obtain in all, (x+a)^v =a^v [ ((v),(v) )((x/a))^v + ((v),((v−1)) )((x/a))^(v−1) + ((v),((v−2)) )((x/a))^(v−2) +...+ ((v),(2) )((x/a))^2 + ((v),(1) )((x/a)) +1] (x+a)^v =a^v (Σ_(r=0) ^v ((v),(r) )x^r a^(−r) ) (x+a)^v =Σ_(r=0) ^v ((v),(r) )x^r a^(v−r) □ If one approached the proof starting with ′The number of ways of obtaining the term ((a/x))^0 from {1+(a/x)}^v (x≠0) (after writting {x+a}^v =x^v (1+(a/x))^v ) is ((v),(0) )...′ you get the equivalent expression (x+a)^v =Σ_(r=0) ^v ((v),(r) )x^(v−r) a^r .$
	${(x + a)}^{v} = {(a {1 + \frac{x}{a}})}^{v}$ ${(x + a)}^{v} = a^{v} {1 + \frac{x}{a}}^{v}$ $H e r e, I a s s u m e t h a t v \in Z^{⩾} . a \neq 0$ $(Z^{⩾} = {n o n - n e g a t i v e i n t e g e r s})$ $T h e c h o o s e f u n c t i o n i s g i v e n a s (\begin{matrix} n \\ r \end{matrix})$ $a s t h e n u m b e r o f w a y s o f c h o o s i n g$ $r o b j e c t s f r o m n d i s t i n c t o b j e c t s .$ $O f c o u r s e w e h a v e$ ${1 + \frac{x}{a}}^{v} = (1 + \frac{x}{a}) (1 + \frac{x}{a}) (1 + \frac{x}{a}) . . . (1 + \frac{x}{a})$ $(v t i m e s)$ $H e r e y o u h a v e v b r a c k e t s w i t h \frac{x}{a} i n$ $e a c h o f t h e m s o t h a t p o w e r s o f \frac{x}{a}$ $c a n a r i s e i n a n u m b e r o f w a y s a s a$ $c o n s e q u e n c e o f p i c k i n g b r a c k e t s t o$ $m u l t i p l y w i t h e a c h o t h e r t o g i v e t h e$ $s o u g h t a f t e r p o w e r o f \frac{x}{a} .$ ${(\frac{x}{a})}^{v} : T o o b t a i n t h i s t e r m, y o u h a v e$ $v b r a c k e t s t o c h o o s e \frac{x}{a} f r o m a n d$ $t h i s i s d o n e i n (\begin{matrix} v \\ v \end{matrix}) w a y s . S o,$ ${(\frac{x}{a})}^{v} o c c u r s (\begin{matrix} v \\ v \end{matrix}) t i m e s a n d t h u s$ $t h e t e r m i n {(\frac{x}{a})}^{v} i s (\begin{matrix} v \\ v \end{matrix}) {(\frac{x}{a})}^{v} .$ ${(\frac{x}{a})}^{v - 1} : H e r e i t i s t h e c a s e t h a t {y o u}^{'} d$ $l i k e t o o b t a i n \frac{x}{a} m u l t i p l i e d$ $b y i t s e l f v - 1 t i m e s .$ $Y o u h a v e v b r a c k e t s t o c h o o s e$ $f r o m, s o t h a t t h i s o c c u r s$ $i n (\begin{matrix} v \\ v - 1 \end{matrix}) w a y s . H e n c e, t h e$ $t e r m i n {(\frac{x}{a})}^{v - 1} i s (\begin{matrix} v \\ v - 1 \end{matrix}) {(\frac{x}{a})}^{v - 1} .$ ${(\frac{x}{a})}^{v - 2} : T h i s t e r m a r i s e s i n (\begin{matrix} v \\ v - 2 \end{matrix}) w a y s$ $f r o m a m o n g t h e b r a c k e t s, s o$ $t h i s t e r m i s (\begin{matrix} v \\ v - 2 \end{matrix}) {(\frac{x}{a})}^{v - 2} .$ $C o n t i n u i n g o n, t e r m s i n {(\frac{x}{a})}^{2}$ $a r i s e i n (\begin{matrix} v \\ 2 \end{matrix}) w a y s, t e r m s i n \frac{x}{a} a r i s e$ $i n (\begin{matrix} v \\ 1 \end{matrix}) w a y s a n d t e r m s i n {(\frac{x}{a})}^{0} a r i s e$ $i n (\begin{matrix} v \\ 0 \end{matrix}) = 1 w a y s f r o m a m o n g t h e b r a c k e t s .$ $W e h e n c e o b t a i n i n a l l,$ ${(x + a)}^{v} = a^{v} [(\begin{matrix} v \\ v \end{matrix}) {(\frac{x}{a})}^{v} + (\begin{matrix} v \\ v - 1 \end{matrix}) {(\frac{x}{a})}^{v - 1}$ $+ (\begin{matrix} v \\ v - 2 \end{matrix}) {(\frac{x}{a})}^{v - 2} + . . . + (\begin{matrix} v \\ 2 \end{matrix}) {(\frac{x}{a})}^{2} + (\begin{matrix} v \\ 1 \end{matrix}) (\frac{x}{a})$ $+ 1]$ ${(x + a)}^{v} = a^{v} (\underset{r = 0}{\sum^{v}} (\begin{matrix} v \\ r \end{matrix}) x^{r} a^{- r})$ ${(x + a)}^{v} = \underset{r = 0}{\sum^{v}} (\begin{matrix} v \\ r \end{matrix}) x^{r} a^{v - r} ◻$ $I f o n e a p p r o a c h e d t h e p r o o f s t a r t i n g$ $w i t h^{'} T h e n u m b e r o f w a y s o f o b t a i n i n g$ $t h e t e r m {(\frac{a}{x})}^{0} f r o m {1 + \frac{a}{x}}^{v} (x \neq 0)$ $(a f t e r w r i t t i n g {x + a}^{v} = x^{v} {(1 + \frac{a}{x})}^{v})$ $i s (\begin{matrix} v \\ 0 \end{matrix}) . . .^{'} y o u g e t t h e e q u i v a l e n t$ $e x p r e s s i o n$ ${(x + a)}^{v} = \underset{r = 0}{\sum^{v}} (\begin{matrix} v \\ r \end{matrix}) x^{v - r} a^{r} .$
	Answered by prakash jain last updated on 13/Dec/15

	$f (x) = {(x + a)}^{n}$ $f^{'} (x) = n {(x + a)}^{n - 1}$ $f^{(2)} (x) = n (n - 1) {(x + a)}^{n - 2}$ $f^{(n)} (x) = n!$ $f^{(k)} (x) = 0, k > n$ $Taylor series$ $f (x) = a^{n} + \frac{n}{1!} x + \frac{n (n - 1)}{2!} x^{2} + . . . + \frac{n!}{n!} x^{n} + 0 x^{n + 1} + . .$ ${(x + a)}^{n} = \underset{i = 0}{\sum^{n}}^{n} C_{i} x^{i} a^{n - i}$
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