Question Number 145557 by puissant last updated on 06/Jul/21
Answered by Olaf_Thorendsen last updated on 06/Jul/21
$$\left.\mathrm{1}\right) \\ $$$$\mathrm{R}_{{n}} \left({x}\right)\:=\:\left(\mathrm{1}+{x}\right)^{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{C}_{{n}} ^{{k}} {x}^{{k}} \\ $$$$\mathrm{changeons}\:\mathrm{d}'\mathrm{indices}\:{et}\:{de}\:{notations}. \\ $$$${n}\:{devient}\:{p}+{q}\:{et}\:{k}\:{devient}\:{n}. \\ $$$$\mathrm{R}_{{p}+{q}} \left({x}\right)\:=\:\left(\mathrm{1}+{x}\right)^{{p}+{q}} \:=\:\underset{{n}=\mathrm{0}} {\overset{{p}+{q}} {\sum}}\mathrm{C}_{{p}+{q}} ^{{n}} {x}^{{n}} \\ $$$$\mathrm{Le}\:\mathrm{coefficient}\:\mathrm{de}\:{x}^{{n}} \:{est}\:{donc}\:\mathrm{C}_{{p}+{q}} ^{{n}} \\ $$$$\left.\mathrm{2}\right) \\ $$$${Mais}\:{x}^{{n}} \:=\:{x}^{{k}} {x}^{{n}−{k}} \\ $$$${En}\:{outre}\:\left(\mathrm{1}+{x}\right)^{{p}+{q}} \:=\:\left(\mathrm{1}+{x}\right)^{{p}} \left(\mathrm{1}+{x}\right)^{{q}} \\ $$$${Le}\:{coefficient}\:{de}\:{x}^{{n}} \:{est}\:{donc}\:{la}\:{somme} \\ $$$${des}\:{produits}\:{des}\:{coefficients}\:{des}\:{x}^{{k}} \\ $$$${par}\:{les}\:{coefficients}\:{des}\:{x}^{{n}−{k}} \:{dans} \\ $$$$\left(\mathrm{1}+{x}\right)^{{p}} \:{et}\:{dans}\:\left(\mathrm{1}+{x}\right)^{{q}} . \\ $$$${Et}\:{donc}\:\mathrm{C}_{{p}+{q}} ^{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{C}_{{p}} ^{{k}} \mathrm{C}_{{q}} ^{{n}−{k}} \\ $$
Commented by puissant last updated on 06/Jul/21
$$\mathrm{merci} \\ $$