Question

Stie cineva cum se rezolva ?
(n+1)! = 5n! + 12(n-1)!

Answer 1

$\displaystyle \mathtt{(n+1)!=5n!+12(n-1)!}\\ \\ \mathtt{n!(n+1)=5 n (n-1)!+12(n-1)!}\\ \\ \mathtt{n(n-1)!(n+1)=5n(n-1)!+12(n-1)!}\\ \\ \mathtt{n(n+1)=5n+12}\\ \\ \mathtt{n^2+n-5n-12=0}\\ \\ \mathtt{n^2-4n-12=0}\\ \\ \mathtt{a=1,~b=-4,~c=-12}\\ \\ \mathtt{\Delta=b^2-4ac=(-4)^2-4 \cdot 1 \cdot (-12)=16+48=64\ \textgreater \ 0}$

$\displaystyle \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-4)+ \sqrt{64} }{2 \cdot 1}= \frac{4+8}{2}= \frac{12}{2}=6 \in \mathbb{N}}\\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-(-4)- \sqrt{64} }{2 \cdot 1} = \frac{4-8}{2}= \frac{-4}{2}=-2 \not \in \mathbb{N} }\\ \\ \mathtt{S=\{6\}}$