Question

Sa se determine numarul real $n \geq 1$, daca
$\frac{(n+1)!}{(n-1)!} =-4n+6$

Answer 1

$\frac{(n+1)!}{(n-1)!}= \frac{1*2*3*...*(n-1)*n*(n+1)}{1*2*3*..*(n-1)}=n(n+1)$

[tex]n(n+1)=-4n+6\\ n^2+n+4n-6=0\\ n^2+5n-6=0\\ \Delta=25 + 4*6=49\\ n_{1,2}= \frac{-5\pm \sqrt{\Delta} }{2}= \frac{-5\pm7}{2} \\ n_1=1\\ n_2=-6[/tex]

NI se spune deja ca n ≥ 1 ==> n nu poate fi -6

Raspuns final n = 1