Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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!

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

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

y′re welcom

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com