Question

Ajutor

a) (n+2)! supra n! =12
b) (n+1)! supra 12 =(n+1)!

Answer 1

a) $\frac{(n+2)!}{n!}=12=> \frac{(n+2)!}{(n+2)!(n+1)n}=12 => \frac{1}{n^{2}+n}=12=> 12n^{2}+12n=1 => 12n^{2}+12n-1=0 \\ delta= b^{2}-4ac=144+48=192 >0\\x1=\frac{-b+\sqrt{delta}} {2a} =\frac{-12+8\sqrt{3}}{24} =\frac{4(-3+2\sqrt{3})}{24}= \frac{-3+2\sqrt{3}} {6}\\x2=\frac{-3-2\sqrt{3}} {6}$

b) nu stiu scuze