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Question-186002




Question Number 186002 by Shrinava last updated on 30/Jan/23
Commented by JDamian last updated on 30/Jan/23
why did you re-post the same question?
Answered by mr W last updated on 30/Jan/23
let′s find the coefficient of x^n  term in  (1+x)^(2n) , which is clearly C_n ^(2n) .   on the other side  (1+x)^(2n) =(1+x)^n (1+x)^n   the coef. of x^n  from (1+x)^n (1+x)^n  is  Σ_(k=0) ^n C_k ^n C_(n−k) ^n .  since C_(n−k) ^n =C_k ^n , Σ_(k=0) ^n C_k ^n C_(n−k) ^n =Σ_(k=0) ^n (C_k ^n )^2 ,  therefore  Σ_(k=0) ^n (C_k ^n )^2 =C_n ^(2n)  ✓
$${let}'{s}\:{find}\:{the}\:{coefficient}\:{of}\:{x}^{{n}} \:{term}\:{in} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} ,\:{which}\:{is}\:{clearly}\:{C}_{{n}} ^{\mathrm{2}{n}} .\: \\ $$$${on}\:{the}\:{other}\:{side} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} =\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${the}\:{coef}.\:{of}\:{x}^{{n}} \:{from}\:\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} \:{is} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {C}_{{n}−{k}} ^{{n}} . \\ $$$${since}\:{C}_{{n}−{k}} ^{{n}} ={C}_{{k}} ^{{n}} ,\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {C}_{{n}−{k}} ^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} , \\ $$$${therefore} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} ={C}_{{n}} ^{\mathrm{2}{n}} \:\checkmark \\ $$
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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