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Question Number 145557 by puissant last updated on 06/Jul/21

Answered by Olaf_Thorendsen last updated on 06/Jul/21

$$1)$$$${\mathrm{R}}_n\left(x\right)={(1+x)}^n=\underset{k=0}{\sum ^n}{\mathrm{C}}_n^k{x}^k$$$$\mathrm{changeons}{\mathrm{d}}^{\prime }\mathrm{indices}etdenotations.$$$$ndevientp+qetkdevientn.$$$${\mathrm{R}}_{p+q}\left(x\right)={(1+x)}^{p+q}=\underset{n=0}{\sum ^{p+q}}{\mathrm{C}}_{p+q}^n{x}^n$$$$\mathrm{Le}\mathrm{coefficient}\mathrm{de}{x}^{n}estdonc{\mathrm{C}}_{p+q}^n$$$$2)$$$$Mais{x}^n=x^k{x}^{n-k}$$$$Enoutre{(1+x)}^{p+q}={(1+x)}^p{(1+x)}^q$$$$Lecoefficientde{x}^{n}estdonclasomme$$$$desproduitsdescoefficientsdes{x}^k$$$$parlescoefficientsdes{x}^{n-k}dans$$$${(1+x)}^{p}etdans{(1+x)}^q.$$$$Etdonc{\mathrm{C}}_{p+q}^n=\underset{k=0}{\sum ^n}{\mathrm{C}}_p^k{\mathrm{C}}_q^{n-k}$$

Commented by puissant last updated on 06/Jul/21

$$\mathrm{merci}$$

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