prove that ∀(n,p)∈N^★ ×N 1)Σ_(k=0) ^p (−1)^k C_n ^k =(−1)^p C_(n−1) ^p 2)∀(p,q)∈N^2 Σ_(k=0) ^p C_(p+q) ^k C_(p+q−k) ^(p−k) =2^p C


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Question Number 73040 by mathmax by abdo last updated on 05/Nov/19

$p r o v e t h a t \forall (n, p) \in N^{★} \times N$ $1) \sum_{k = 0}^{p} {(- 1)}^{k} C_{n}^{k} = {(- 1)}^{p} C_{n - 1}^{p}$ $2) \forall (p, q) \in N^{2} \sum_{k = 0}^{p} C_{p + q}^{k} C_{p + q - k}^{p - k} = 2^{p} C_{p + q}^{p}$
	Answered by mind is power last updated on 06/Nov/19
	$1) recursion 2 nd lets A=[1,,,p+q] in A we can get C_(p+q) ^p set withe p elements the number of idont now the nam in inglish ′′A={1,2}/P(A)={∅,{1},{2},{1,2}}′′ if A is card p⇒P(A)=2^p sl the number all partition of set withe p element is 2^p C_(p+q) ^p we can chose this element we pick k element in p+q and p−k in p+q−k ⇒=Σ_(k=0) ^p C_(p+q) ^k C_(p+q−k) ^(p−k) =2^p C_(p+q) ^p$
	$1) recursion$ $2 nd$ $lets A = [1,,, p + q]$ $in A we can get$ $C_{p + q}^{p} set withe p elements$ $the number of idont now the nam in inglish$ $^{″} A = {1, 2} / P (A) = {\emptyset, {1}, {2}, {1, 2}}^{″}$ $if A is card p \Rightarrow P (A) = 2^{p}$ $sl the number all partition of set withe p element is 2^{p} C_{p + q}^{p}$ $we can chose this element$ $we pick k element in p + q$ $and p - k in p + q - k$ $\Rightarrow= \underset{k = 0}{\sum^{p}} C_{p + q}^{k} C_{p + q - k}^{p - k} = 2^{p} C_{p + q}^{p}$
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