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
$${let}'{s}\:{find}\:{the}\:{coefficient}\:{of}\:{x}^{{n}} \:{term}\:{in} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} ,\:{which}\:{is}\:{clearly}\:{C}_{{n}} ^{\mathrm{2}{n}} .\: \\ $$$${on}\:{the}\:{other}\:{side} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}} =\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} \\ $$$${the}\:{coef}.\:{of}\:{x}^{{n}} \:{from}\:\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+{x}\right)^{{n}} \:{is} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {C}_{{n}−{k}} ^{{n}} . \\ $$$${since}\:{C}_{{n}−{k}} ^{{n}} ={C}_{{k}} ^{{n}} ,\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {C}_{{n}−{k}} ^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} , \\ $$$${therefore} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{k}} ^{{n}} \right)^{\mathrm{2}} ={C}_{{n}} ^{\mathrm{2}{n}} \:\checkmark \\ $$
Commented by Shrinava last updated on 30/Jan/23
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you} \\ $$