Question-198231

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

Answered by witcher3 last updated on 14/Oct/23

C_{n}^{p} = \frac{n!}{p! . (n - p)!}

{pC}_{n}^{p} = \frac{n!}{(p - 1)! . (n - p)!} = n . \frac{(n - 1)!}{(p - 1)! . (n - 1 - (p - 1))!}

= {nC}_{n - 1}^{p - 1} . . \forall p ⩾ 1

\Leftrightarrow \frac{1}{p} C_{n - 1}^{p - 1} = \frac{C_{n}^{p}}{n}

S_{n} = \underset{k = 0}{\sum^{n}} \frac{C_{n + 1 - 1}^{k + 1 - 1}}{k + 1} = \underset{k = 0}{\sum^{n}} \frac{C_{n + 1}^{k + 1}}{n + 1} = \frac{1}{n + 1} \underset{k = 1}{\sum^{n + 1}} C_{n + 1}^{k}

= \frac{1}{n + 1} (2^{n + 1} - 1)

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