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Question Number 73041 by mathmax by abdo last updated on 05/Nov/19

prove that ∀n∈ N Σ_(k=0) ^(2n) (−1)^k (C_(2n) ^k )^2 =(−1)^n C_(2n) ^n

$${prove}\:{that}\:\forall{n}\in\:{N}\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:\left(−\mathrm{1}\right)^{{k}} \:\left({C}_{\mathrm{2}{n}} ^{{k}} \right)^{\mathrm{2}} \:=\left(−\mathrm{1}\right)^{{n}} \:{C}_{\mathrm{2}{n}} ^{{n}} \\ $$

Commented by mathmax by abdo last updated on 06/Nov/19

let p(x)=(1+x)^(2n) and q(x)=(1−x)^(2n) ⇒ p(x).q(x)=(Σ_(k=0) ^(2n) C_(2n) ^k x^k )(Σ_(k=0) ^(2n) C_(2n) ^k (−1)^k x^k ) =Σ_(k=0) ^(2n) C_k x^k with C_k =Σ_(i+j=k) a_i b_j =Σ_(i=0) ^k a_i b_(k−i) =Σ_(i=0) ^k C_(2n) ^i (−1)^(k−i) C_(2n) ^(k−i) ⇒the coefficient of x^(2n) is C_(2n) =Σ_(i=0) ^(2n) C_(2n) ^i (−1)^(2n−i) C_(2n) ^(2n−i) =Σ_(i=0) ^(2n) (−1)^i (C_(2n) ^i )^2 also we have p(x)q(x)=(1−x^2 )^(2n) =Σ_(k=0) ^(2n) (−1)^k C_(2n) ^k x^(2k) ⇒the coefficient of x^(2n) is is (−1)^n C_(2n) ^n ⇒ Σ_(i=0) ^(2n) (−1)^i (C_(2n) ^i )^2 =(−1)^n C_(2n) ^n

$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \:{and}\:{q}\left({x}\right)=\left(\mathrm{1}−{x}\right)^{\mathrm{2}{n}} \:\Rightarrow \\ $$$${p}\left({x}\right).{q}\left({x}\right)=\left(\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \:{x}^{{k}} \right)\left(\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} \:{x}^{{k}} \right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{{k}} {x}^{{k}} \:{with}\:{C}_{{k}} =\sum_{{i}+{j}={k}} {a}_{{i}} {b}_{{j}} =\sum_{{i}=\mathrm{0}} ^{{k}} \:{a}_{{i}} {b}_{{k}−{i}} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{k}} {C}_{\mathrm{2}{n}} ^{{i}} \left(−\mathrm{1}\right)^{{k}−{i}} \:{C}_{\mathrm{2}{n}} ^{{k}−{i}} \:\:\Rightarrow{the}\:{coefficient}\:{of}\:{x}^{\mathrm{2}{n}} \:{is} \\ $$$${C}_{\mathrm{2}{n}} =\sum_{{i}=\mathrm{0}} ^{\mathrm{2}{n}} \:{C}_{\mathrm{2}{n}} ^{{i}} \left(−\mathrm{1}\right)^{\mathrm{2}{n}−{i}} \:{C}_{\mathrm{2}{n}} ^{\mathrm{2}{n}−{i}} \:=\sum_{{i}=\mathrm{0}} ^{\mathrm{2}{n}} \left(−\mathrm{1}\right)^{{i}} \left({C}_{\mathrm{2}{n}} ^{{i}} \right)^{\mathrm{2}} \:{also}\:{we}\:{have} \\ $$$${p}\left({x}\right){q}\left({x}\right)=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{n}} \:=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \left(−\mathrm{1}\right)^{{k}} {C}_{\mathrm{2}{n}} ^{{k}} \:{x}^{\mathrm{2}{k}} \:\:\Rightarrow{the}\:{coefficient}\:{of}\:{x}^{\mathrm{2}{n}} {is} \\ $$$${is}\:\left(−\mathrm{1}\right)^{{n}} \:{C}_{\mathrm{2}{n}} ^{{n}} \:\Rightarrow\:\sum_{{i}=\mathrm{0}} ^{\mathrm{2}{n}} \:\left(−\mathrm{1}\right)^{{i}} \left({C}_{\mathrm{2}{n}} ^{{i}} \right)^{\mathrm{2}} \:=\left(−\mathrm{1}\right)^{{n}} \:{C}_{\mathrm{2}{n}} ^{{n}} \\ $$

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

let p(x)=(1+x)^(2n) Q(x)=(1−x)^(2n) p(x)=Σ_(k=0) ^n C_(2n) ^k X^k Q(x)=Σ_(k=0) ^n C_n ^k (−x)^k lets find coeficient of X^(2n) in p.Q P,Q=Σ_(k=0) ^(2n) C_(2n) ^k (−1)^k x^k .Σ_(j=0) ^(2n) C_(2n) ^j x^j ⇒k+j=2n =Σ_(k=0) ^(2n) (−1)^k C_(2n) ^k .C_(2n) ^(2n−k) =Σ_(k=0) ^(2n) (−1)^k (C_(2n) ^k )^2 p.Q=(1−x^2 )^(2n) coeficient of X^(2n) isC_(2n) ^n .(−1)^n ⇒Σ_(k=0) ^(2n) (−1)^k (C_(2n) ^k )^2 =(−1)^n .C_(2n) ^n

$$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2n}} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2n}} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{2n}} ^{\mathrm{k}} \mathrm{X}^{\mathrm{k}} \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \left(−\mathrm{x}\right)^{\mathrm{k}} \\ $$$$\mathrm{lets}\:\mathrm{find}\:\mathrm{coeficient}\:\mathrm{of}\:\mathrm{X}^{\mathrm{2n}} \:\mathrm{in}\:\mathrm{p}.\mathrm{Q} \\ $$$$\mathrm{P},\mathrm{Q}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\mathrm{C}_{\mathrm{2n}} ^{\mathrm{k}} \left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{x}^{\mathrm{k}} .\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\mathrm{C}_{\mathrm{2n}} ^{\mathrm{j}} \mathrm{x}^{\mathrm{j}} \\ $$$$\Rightarrow\mathrm{k}+\mathrm{j}=\mathrm{2n} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{C}_{\mathrm{2n}} ^{\mathrm{k}} .\mathrm{C}_{\mathrm{2n}} ^{\mathrm{2n}−\mathrm{k}} =\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \left(\mathrm{C}_{\mathrm{2n}} ^{\mathrm{k}} \right)^{\mathrm{2}} \\ $$$$\mathrm{p}.\mathrm{Q}=\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2n}} \\ $$$$\mathrm{coeficient}\:\mathrm{of}\:\mathrm{X}^{\mathrm{2n}} \mathrm{isC}_{\mathrm{2n}} ^{\mathrm{n}} .\left(−\mathrm{1}\right)^{\mathrm{n}} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \left(\mathrm{C}_{\mathrm{2n}} ^{\mathrm{k}} \right)^{\mathrm{2}} =\left(−\mathrm{1}\right)^{\mathrm{n}} .\mathrm{C}_{\mathrm{2n}} ^{\mathrm{n}} \\ $$

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