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Aratati ca pt n∈N

√1+2+...+n/√1+2+...+(n+1)×√n+2/√n ∈N



Răspuns :

[tex]\frac{ \sqrt{1+2+...+n} }{ \sqrt{1+2+...+(n+1)} } * \frac{ \sqrt{n+2} }{ \sqrt{n} }= \\ = \frac{ \sqrt{ \frac{n(n+1)}{2} } }{ \sqrt{ \frac{(n+1)(n+2)}{2} } } * \frac{ \sqrt{n+2} }{ \sqrt{n} }=\\ = \frac{ \sqrt{ n(n+1)} }{ \sqrt{ (n+1)(n+2) } } * \frac{ \sqrt{n+2} }{ \sqrt{n} }=1\in N[/tex]