Question

exercițiul 4 ma poate ajuta cineva?​

Answer 1

 

$\displaystyle\bf\\1+C_{n+1}^1-n^2=0\\\\1+(n+1)-n^2=0\\\\-n^2+n+2=0~~\Big|~~\cdot(-1)\\\\n^2-n-2=0\\\\n_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{1\pm\sqrt{1+8}}{2}=\frac{1\pm\sqrt{9}}{2}=\frac{1\pm3}{2}\\\\n_1=\frac{1+3}{2}=\frac{4}{2}=2\\\\n_2=\frac{1-3}{2}=\frac{-2}{2}=-1 \notin N~~(Solutie~eliminata.)\\\\\implies~~\boxed{\bf~n=2}$