Question

Aratati ca : { $\sqrt{ n^{2}+4n+5 }$}<$\frac{1}{2}$ pentru orice n apartinand lui N*

Answer 1

$\displaystyle n^2+4n+5=(n+2)^2+1\ \textgreater \ (n+2)^2 \\ \\ n^2+4n+5\ \textless \ n^2+6n+9=(n+3)^2 \\ \\ Deci~n+2\ \textless \ \sqrt{n^2+4n+5}\ \textless \ n+3,~ceea~ce~inseamna~ca \\ \\ \left [ \sqrt{n^2+4n+5 \right]=n+2. \\ \\ Atunci~ \left \{ \sqrt{n^2+4n+5} \right \}= \sqrt{n^2+4n+5}-\left [ \sqrt{n^2+4n+5 \right]= \\ \\ = \sqrt{n^2+4n+5}-n-2.$

$\displaystyle Deci~trebuie~sa~demonstram~ca~\sqrt{n^2+4n+5}-n-2\ \textless \ \frac{1}{2} \Leftrightarrow \\ \\ \Leftrightarrow \sqrt{n^2+4n+5}\ \textless \ n+ \frac{5}{2} \Leftrightarrow n^2+4n+5\ \textless \ n^2+5n+ \frac{25}{4} \Leftrightarrow \\ \\ \Leftrightarrow 5-\frac{25}{4}\ \textless \ n,~adevarat!$