Question

Exercitiul este in atasament !!!

Answer 1

$\displaystyle Din~C^3_n~|~C^3_{n+1}~si~C^3_n,~C^3_{n+1}>0~rezulta~ \frac{C^3_{n+1}}{C^3_n} \in \mathbb{N^*}. \\ \\ C^3_n= \frac{n!}{(n-3)! \cdot 3!}= \frac{(n-2)(n-1)n}{6}. \\ \\ C^3_{n+1}= \frac{(n-1)n(n+1)}{6}. \\ \\ \frac{C^3_{n+1}}{C^3_n}= \frac{n+1}{n-2}=1+ \frac{3}{n-2} \in \mathbb{N^*} \Rightarrow \frac{3}{n-2} \in \mathbb{N} \Rightarrow n-2 \in \{1,3\} \Rightarrow \\ \\ \Rightarrow n \in \{3,5\}.$

Answer 2

Am atasat o rezolvare.