Question

Determinati n aparține lui N, n > sau = 2, pentru care C n luate cate 1 + C n luate cate 2 = 15.

Answer 1

$\displaystyle C_n^1+C_n^2=15~~~~~~~~~~~~~~~~~~~~~~~~n \geq 2 \Rightarrow n \in [2, \infty) \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{C_n^1=n} \\ n+ \frac{n!}{(n-2)! \cdot 2!}=15 \\ \\ n+ \frac{(n-2)!(n-1) \cdot n}{(n-2)! \cdot 2} =15 \\ \\ n+ \frac{(n-1) \cdot n}{2} =15 \\ \\ 2n+(n-1) \cdot n=15 \cdot 2 \\ 2n+n^2-n=30 \\ n^2+n-30=0 \\ a=1,~b=1,~c=-30$
$\displaystyle \Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-30)= 1+ 120=121\ \textgreater \ 0 \\ n_1= \frac{-1- \sqrt{121} }{2 \cdot 1} = \frac{-1-11}{2} = \frac{-12}{2} -6 \\ \\ n_2= \frac{-1+ \sqrt{121} }{2 \cdot 1} = \frac{-1+11}{2} = \frac{10}{2} =5 \\ \\ \left\begin{array}{cc}n_1=-6 \not \in [2, \infty) \\n_2=5 \in [2, \infty)\\\end{array}\right\} \Rightarrow S=\{5\}$