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Question-185982




Question Number 185982 by Shrinava last updated on 30/Jan/23
Answered by Mathspace last updated on 30/Jan/23
we developp first (x+1)^(2n)   =Σ_(k=0) ^(2n) C_(2n) ^k x^k   but  (x+1)^(2n) =(x+1)^n .(x+1)^n   =(Σ_(k=0) ^n C_n ^k x^k )(Σ_(k=0) ^n C_n ^k x^k )  =Σ_(k=0) ^(2n)  a_k x^k   a_k =Σ_(i+j=k) a_i a_j   =Σ_(i=0) ^k a_i a_(k−i) =Σ_(i=0) ^k C_k ^i C_k ^(k−i)   =Σ_(i=0) ^k (C_k ^i )^2   the equality give  Σ_(i=0) ^k (C_k ^i )^2 =C_(2n) ^k   k=n ⇒Σ_(i=0) ^n (C_n ^i )^2 =C_(2n) ^n   =(((2n)!)/((n)!^2 ))
$${we}\:{developp}\:{first}\:\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} {C}_{\mathrm{2}{n}} ^{{k}} {x}^{{k}} \:\:{but} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} =\left({x}+\mathrm{1}\right)^{{n}} .\left({x}+\mathrm{1}\right)^{{n}} \\ $$$$=\left(\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} {x}^{{k}} \right)\left(\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} {x}^{{k}} \right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:{a}_{{k}} {x}^{{k}} \\ $$$${a}_{{k}} =\sum_{{i}+{j}={k}} {a}_{{i}} {a}_{{j}} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{k}} {a}_{{i}} {a}_{{k}−{i}} =\sum_{{i}=\mathrm{0}} ^{{k}} {C}_{{k}} ^{{i}} {C}_{{k}} ^{{k}−{i}} \\ $$$$=\sum_{{i}=\mathrm{0}} ^{{k}} \left({C}_{{k}} ^{{i}} \right)^{\mathrm{2}} \\ $$$${the}\:{equality}\:{give} \\ $$$$\sum_{{i}=\mathrm{0}} ^{{k}} \left({C}_{{k}} ^{{i}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{k}} \\ $$$${k}={n}\:\Rightarrow\sum_{{i}=\mathrm{0}} ^{{n}} \left({C}_{{n}} ^{{i}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$$$=\frac{\left(\mathrm{2}{n}\right)!}{\left({n}\right)!^{\mathrm{2}} } \\ $$
Commented by Shrinava last updated on 30/Jan/23
cool dear professor thank you
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$

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