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Question Number 198231 by pticantor last updated on 14/Oct/23

Answered by witcher3 last updated on 14/Oct/23

$${\mathrm{C}}_{\mathrm{n}}^{\mathrm{p}}=\frac{\mathrm{n}!}{\mathrm{p}!.(\mathrm{n}-\mathrm{p})!}$$$${\mathrm{pC}}_{\mathrm{n}}^{\mathrm{p}}=\frac{\mathrm{n}!}{(\mathrm{p}-1)!.(\mathrm{n}-\mathrm{p})!}=\mathrm{n}.\frac{(\mathrm{n}-1)!}{(\mathrm{p}-1)!.(\mathrm{n}-1-(\mathrm{p}-1))!}$$$$={\mathrm{nC}}_{\mathrm{n}-1}^{\mathrm{p}-1}..\mathrm{\forall }\mathrm{p}⩾1$$$$\iff \frac{1}{\mathrm{p}}{\mathrm{C}}_{\mathrm{n}-1}^{\mathrm{p}-1}=\frac{{\mathrm{C}}_{\mathrm{n}}^{\mathrm{p}}}{\mathrm{n}}$$$${\mathrm{S}}_{\mathrm{n}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{{\mathrm{C}}_{\mathrm{n}+1-1}^{\mathrm{k}+1-1}}{\mathrm{k}+1}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\frac{{\mathrm{C}}_{\mathrm{n}+1}^{\mathrm{k}+1}}{\mathrm{n}+1}=\frac{1}{\mathrm{n}+1}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}+1}}{\mathrm{C}}_{\mathrm{n}+1}^{\mathrm{k}}$$$$=\frac{1}{\mathrm{n}+1}(2^{\mathrm{n}+1}-1)$$

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