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Question Number 26133 by abdo imad last updated on 21/Dec/17
find the value of  (C_n ^(0  ) )^2  +(C_n ^1 )^2  +(C_n ^2 )^2  +...(C_n ^n )^2 .
$${find}\:{the}\:{value}\:{of}\:\:\left({C}_{{n}} ^{\mathrm{0}\:\:} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{1}} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{2}} \right)^{\mathrm{2}} \:+…\left({C}_{{n}} ^{{n}} \right)^{\mathrm{2}} . \\ $$
Commented by abdo imad last updated on 21/Dec/17
let put  A_n =( C_n ^0  )^2   +( C_n ^1  )^2   +...+(C_n ^n  )^2   we have  (1+x)^(2n)  =  Σ_(k=0) ^(k=2n)  x^k   = C_(2n) ^0 +C_(2n) ^1 x +C_(2n) ^2 x^2  +...C_(2n) ^(2n)  x^(2n)   but (1+x)^(2n) =(1+x)^n .(1+x)^n =( Σ_(k=0) ^(k=n)  C_n ^k  x^k )(Σ_(k=0) ^(k=n)   C_n ^k  x^k )  = Σ_(k=0) ^(k=2n)  λ_k  x^k    and  λ_k =Σ_(i+j=k)  a_i  b_j   =  Σ_(i=0) ^(i=k)   a_i  b_(k−i)   =  Σ_(i=0) ^(i=k)   C_n ^i  C_n ^(k−i)   −−>λ_n = Σ_(i=0) ^(i=n)  C_n ^i   C_n ^(n−i)   =Σ_(i=0) ^(i=n)  ( C_n ^i )^2   but λ_n  is the coefficient of x^n −−> λ_n =C_(2n) ^n   A_n = C_(2n) ^n   =  (((2n)!)/((n!)^2 ))  .
$${let}\:{put}\:\:{A}_{{n}} =\left(\:{C}_{{n}} ^{\mathrm{0}} \:\right)^{\mathrm{2}} \:\:+\left(\:{C}_{{n}} ^{\mathrm{1}} \:\right)^{\mathrm{2}} \:\:+…+\left({C}_{{n}} ^{{n}} \:\right)^{\mathrm{2}} \\ $$$${we}\:{have}\:\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} \:=\:\:\sum_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{n}} \:{x}^{{k}} \:\:=\:{C}_{\mathrm{2}{n}} ^{\mathrm{0}} +{C}_{\mathrm{2}{n}} ^{\mathrm{1}} {x}\:+{C}_{\mathrm{2}{n}} ^{\mathrm{2}} {x}^{\mathrm{2}} \:+…{C}_{\mathrm{2}{n}} ^{\mathrm{2}{n}} \:{x}^{\mathrm{2}{n}} \\ $$$${but}\:\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} =\left(\mathrm{1}+{x}\right)^{{n}} .\left(\mathrm{1}+{x}\right)^{{n}} =\left(\:\sum_{{k}=\mathrm{0}} ^{{k}={n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \right)\left(\sum_{{k}=\mathrm{0}} ^{{k}={n}} \:\:{C}_{{n}} ^{{k}} \:{x}^{{k}} \right) \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{n}} \:\lambda_{{k}} \:{x}^{{k}} \:\:\:{and}\:\:\lambda_{{k}} =\sum_{{i}+{j}={k}} \:{a}_{{i}} \:{b}_{{j}} \\ $$$$=\:\:\sum_{{i}=\mathrm{0}} ^{{i}={k}} \:\:{a}_{{i}} \:{b}_{{k}−{i}} \:\:=\:\:\sum_{{i}=\mathrm{0}} ^{{i}={k}} \:\:{C}_{{n}} ^{{i}} \:{C}_{{n}} ^{{k}−{i}} \\ $$$$−−>\lambda_{{n}} =\:\sum_{{i}=\mathrm{0}} ^{{i}={n}} \:{C}_{{n}} ^{{i}} \:\:{C}_{{n}} ^{{n}−{i}} \:\:=\sum_{{i}=\mathrm{0}} ^{{i}={n}} \:\left(\:{C}_{{n}} ^{{i}} \right)^{\mathrm{2}} \\ $$$${but}\:\lambda_{{n}} \:{is}\:{the}\:{coefficient}\:{of}\:{x}^{{n}} −−>\:\lambda_{{n}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$$${A}_{{n}} =\:{C}_{\mathrm{2}{n}} ^{{n}} \:\:=\:\:\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }\:\:. \\ $$
Answered by Tinkutara last updated on 21/Dec/17
Commented by abdo imad last updated on 21/Dec/17
a lots of calculus but the method is true .you can  use Σ its more easy...thanks....
$${a}\:{lots}\:{of}\:{calculus}\:{but}\:{the}\:{method}\:{is}\:{true}\:.{you}\:{can} \\ $$$${use}\:\Sigma\:{its}\:{more}\:{easy}…{thanks}…. \\ $$

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