prove-that-for-n-p-N-2-k-0-p-k-C-n-p-k-C-n-k-n-C-2n-1-p-1-conclude-the-value-of-k-0-n-k-C-n-k-2-

Question Number 73042 by mathmax by abdo last updated on 05/Nov/19

prove that for (n,p)∈N^★^2 Σ_(k=0) ^(p ) k C_n ^(p−k) C_n ^k =n C_(2n−1) ^(p−1) conclude the value of Σ_(k=0) ^n k (C_n ^k )^2

$${prove}\:{that}\:{for}\:\left({n},{p}\right)\in{N}^{\bigstar^{\mathrm{2}} } \:\:\:\sum_{{k}=\mathrm{0}} ^{{p}\:} \:{k}\:{C}_{{n}} ^{{p}−{k}} \:{C}_{{n}} ^{{k}} \:={n}\:{C}_{\mathrm{2}{n}−\mathrm{1}} ^{{p}−\mathrm{1}} \\ $$$${conclude}\:{the}\:{value}\:{of}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{k}\:\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} \\ $$

Answered by mind is power last updated on 05/Nov/19

let p(x)=(x+1)^n =Σ_(k=0) ^n C_n ^k X^k p(x).p′(x)=Σ_(k=0) ^n Σ_(j=1) ^n C_k ^n C_j ^n x^k .jx^(j−1) coeficient of power p−1 is when k+j=p =Σ_(k=0) ^n jC_n ^(p−j) C_n ^j X^(p−1) p^2 =(x+1)^(2n) ⇒p^2 ′=2n(x+1)^(2n−1) =2p′.p⇒pp′=n(x+1)^(2n−1) coeficent of x^(p−1) is nC_(2n−1) ^(p−1) ⇒Σ_(k=0) ^p kC_n ^(p−k) C_n ^k =nC_(2n−1) ^(p−1) if p=n ⇒Σk(C_n ^k )^2 =nC_(2n−1) ^(n−1)

$$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \mathrm{X}^{\mathrm{k}} \\ $$$$\mathrm{p}\left(\mathrm{x}\right).\mathrm{p}'\left(\mathrm{x}\right)=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\underset{\mathrm{j}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{k}} ^{\mathrm{n}} \mathrm{C}_{\mathrm{j}} ^{\mathrm{n}} \mathrm{x}^{\mathrm{k}} .\mathrm{jx}^{\mathrm{j}−\mathrm{1}} \\ $$$$\mathrm{coeficient}\:\mathrm{of}\:\mathrm{power}\:\mathrm{p}−\mathrm{1}\:\mathrm{is}\:\mathrm{when}\:\mathrm{k}+\mathrm{j}=\mathrm{p} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{jC}_{\mathrm{n}} ^{\mathrm{p}−\mathrm{j}} \mathrm{C}_{\mathrm{n}} ^{\mathrm{j}} \mathrm{X}^{\mathrm{p}−\mathrm{1}} \\ $$$$\mathrm{p}^{\mathrm{2}} =\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2n}} \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{2}} '=\mathrm{2n}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2n}−\mathrm{1}} =\mathrm{2p}'.\mathrm{p}\Rightarrow\mathrm{pp}'=\mathrm{n}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2n}−\mathrm{1}} \\ $$$$\mathrm{coeficent}\:\mathrm{of}\:\mathrm{x}^{\mathrm{p}−\mathrm{1}} \mathrm{is}\:\mathrm{nC}_{\mathrm{2n}−\mathrm{1}} ^{\mathrm{p}−\mathrm{1}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{p}} {\sum}}\mathrm{kC}_{\mathrm{n}} ^{\mathrm{p}−\mathrm{k}} \mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} =\mathrm{nC}_{\mathrm{2n}−\mathrm{1}} ^{\mathrm{p}−\mathrm{1}} \\ $$$$\mathrm{if}\:\mathrm{p}=\mathrm{n} \\ $$$$\Rightarrow\Sigma\mathrm{k}\left(\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \right)^{\mathrm{2}} =\mathrm{nC}_{\mathrm{2n}−\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \\ $$

Commented by mr W last updated on 06/Nov/19

$${i}\:{learnt}\:{a}\:{new}\:{method}.\:{thanks}\:{sir}! \\ $$

Commented by mind is power last updated on 06/Nov/19

$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom}\: \\ $$

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