Question Number 190857 by AlbertEinstein999 last updated on 13/Apr/23
Commented by mr W last updated on 13/Apr/23
$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} =\mathrm{2}^{{n}} \cancel{−\mathrm{1}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({C}_{{n}} ^{{k}} \right)^{\mathrm{2}} ={C}_{\mathrm{2}{n}} ^{{n}} \\ $$
Commented by mehdee42 last updated on 13/Apr/23
$${why}\:\:\mathrm{2}^{{n}} −\mathrm{1}\:?\:\: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} =\mathrm{2}^{{n}} \\ $$$$ \\ $$
Commented by mr W last updated on 13/Apr/23
$${yes}. \\ $$$${i}\:{had}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{n}} ^{{k}} \:{in}\:{head}. \\ $$