x and y are reals(or complex) let put x^((0)) =1 ,x^((1)) =x x^((2)) =x(x−1).....x^((n)) =x(x−1)(x−2)...(x−n+1)prove that (x+y)^((n)) =Σ_(k=0) ^n C


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Question Number 73027 by mathmax by abdo last updated on 05/Nov/19

$x a n d y a r e r e a l s (o r c o m p l e x) l e t p u t x^{(0)} = 1, x^{(1)} = x$ $x^{(2)} = x (x - 1) . . . . . x^{(n)} = x (x - 1) (x - 2) . . . (x - n + 1) p r o v e t h a t$ ${(x + y)}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} x^{(n - k)} y^{(k)}$
	Answered by mind is power last updated on 05/Nov/19
	$for n=0 (x+y)^0 =1=C_0 ^0 x^0 y^0 =1 suppose ∀n∈N (x+y)^n =ΣC_k ^n x^((n−k)) y^k (x+y)^(n+1) =(x+y)^n .(x+y−n) ={ΣC_n ^k x^((n−k)) y^((k)) }(x+y−n) n=(n−k)+k (x+y−n)=(x−(n−k)+y−k) ⇒Σ_(k=0) ^n C_n ^k x^((n−k)) .y^((k)) .(x−(n−k))+Σ_(k=0) ^n C_n ^k x^((n−k)) y^k .(y−k) =Σ_(k=0) ^n C_n ^k x^((n−k)) .(x−(n−k)).y^k +Σ_(k=0) ^n C_n ^k x^((n−k)) .y^k .(y−k) x^((n−k)) (x−(n−k))=x^((n+1−k)) y^k .(y−k)=y^((k+1)) =Σ_(k=0) ^n C_n ^k x^((n+1−k)) y^k +Σ_(k=0) ^n C_n ^k x^((n−k)) y^(k+1) k→k+1 in 2nd =Σ_(k=0) ^n C_n ^k x^((n+1−k)) y^k +Σ_(k=1) ^(n+1) C_n ^(k−1) x^((n−k+1)) y^((k)) =Σ_(k=1) ^n (C_n ^k +C_n ^k )x^((n+1−k)) y^((k)) +C_0 ^0 x^((n+1)) y^0 +C_n ^n x^0 y^((n+1)) C_n ^n x^0 y^((n+1)) =C_(n+1) ^(n+1) x^((n+1−(n+1))) y^((n+1)) C_n ^0 x^(n+1) y^0 =C_(n+1) ^0 .x^((n+1−0)) y^((0)) C_n ^k +C_n ^(k−1) =((n!)/(k!(n−k)!)).((n!)/((k−1)!.(n−k+1)!))=((n!.(n+1−k)+n!.k)/(k!(n+1−k)!)) =(((n+1)!)/(k!(n+1−k)!))=C_(n+1) ^k =Σ_(k=1) ^n C_(n+1) ^k x^((n+1−k)) y^k +C_(n+1) ^0 x^(n+1) y^0 +C_(n+1) ^(n+1) x^((n+1−(n+1))) y^(n+1) =Σ_(k=0) ^(n+1) x^((n+1−k)) y^((k)) =(x+y)^(n+1)$
	$for n = 0$ ${(x + y)}^{0} = 1 = C_{0}^{0} x^{0} y^{0} = 1$ $suppose \forall n \in N {(x + y)}^{n} = Σ C_{k}^{n} x^{(n - k)} y^{k}$ ${(x + y)}^{n + 1} = {(x + y)}^{n} . (x + y - n)$ $= {Σ C_{n}^{k} x^{(n - k)} y^{(k)}} (x + y - n)$ $n = (n - k) + k$ $(x + y - n) = (x - (n - k) + y - k)$ $\Rightarrow \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{(n - k)} . y^{(k)} . (x - (n - k)) + \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{(n - k)} y^{k} . (y - k)$ $= \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{(n - k)} . (x - (n - k)) . y^{k} + \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{(n - k)} . y^{k} . (y - k)$ $x^{(n - k)} (x - (n - k)) = x^{(n + 1 - k)}$ $y^{k} . (y - k) = y^{(k + 1)}$ $= \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{(n + 1 - k)} y^{k} + \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{(n - k)} y^{k + 1}$ $k \to k + 1 in 2 nd$ $= \underset{k = 0}{\sum^{n}} C_{n}^{k} x^{(n + 1 - k)} y^{k} + \underset{k = 1}{\sum^{n + 1}} C_{n}^{k - 1} x^{(n - k + 1)} y^{(k)}$ $= \underset{k = 1}{\sum^{n}} (C_{n}^{k} + C_{n}^{k}) x^{(n + 1 - k)} y^{(k)} + C_{0}^{0} x^{(n + 1)} y^{0} + C_{n}^{n} x^{0} y^{(n + 1)}$ $C_{n}^{n} x^{0} y^{(n + 1)} = C_{n + 1}^{n + 1} x^{(n + 1 - (n + 1))} y^{(n + 1)}$ $C_{n}^{0} x^{n + 1} y^{0} = C_{n + 1}^{0} . x^{(n + 1 - 0)} y^{(0)}$ $C_{n}^{k} + C_{n}^{k - 1} = \frac{n!}{k! (n - k)!} . \frac{n!}{(k - 1)! . (n - k + 1)!} = \frac{n! . (n + 1 - k) + n! . k}{k! (n + 1 - k)!}$ $= \frac{(n + 1)!}{k! (n + 1 - k)!} = C_{n + 1}^{k}$ $= \underset{k = 1}{\sum^{n}} C_{n + 1}^{k} x^{(n + 1 - k)} y^{k} + C_{n + 1}^{0} x^{n + 1} y^{0} + C_{n + 1}^{n + 1} x^{(n + 1 - (n + 1))} y^{n + 1}$ $= \underset{k = 0}{\sum^{n + 1}} x^{(n + 1 - k)} y^{(k)} = {(x + y)}^{n + 1}$
	Commented by mathmax by abdo last updated on 05/Nov/19

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 06/Nov/19

	$most welcom$
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