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Question-209433




Question Number 209433 by essaad last updated on 10/Jul/24
Answered by A5T last updated on 10/Jul/24
S_p =((1×2×3×4×...×(2p))/(2×4×6×...×2p))=(((2p)!)/(2^p (1×2×...×p)))  =(((2p)!)/(2^p p!))⇒C
$${S}_{{p}} =\frac{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×…×\left(\mathrm{2}{p}\right)}{\mathrm{2}×\mathrm{4}×\mathrm{6}×…×\mathrm{2}{p}}=\frac{\left(\mathrm{2}{p}\right)!}{\mathrm{2}^{{p}} \left(\mathrm{1}×\mathrm{2}×…×{p}\right)} \\ $$$$=\frac{\left(\mathrm{2}{p}\right)!}{\mathrm{2}^{{p}} {p}!}\Rightarrow{C} \\ $$

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