Inductie Matematica

Thibby Vin Dec 14, 2012 2:56 am

Scuze ca nu am scris problema dar sunt in mare graba ... maine dau test din exercitiile astea si nu stiu cum se rezolva ... va rog ajutor .

**Admin** Vin Dec 14, 2012 7:42 pm

Sub-punctul c)

$\begin{array}{l} P\left( k \right):\sum\limits_{k = 1}^n {\sqrt k } \ge \frac{{n\left( {\sqrt n + 1} \right)}}{2}\\ \\ \left. {\color{Red} 1} \right){\rm{Verificare}}\\ \\ P\left( 1 \right):\sqrt 1 = 1 \ge \frac{{1\left( {\sqrt 1 + 1} \right)}}{2} = 1 \to {\rm{Adevarat}}\\ \\ P\left( 2 \right):\sqrt 1 + \sqrt 2 \approx 2.41 \ge \frac{{2\left( {\sqrt 2 + 1} \right)}}{2} = 2.41 \to {\rm{Adevarat}}\\ \\ \left. {\color{Red} 2} \right){\text{Presupunem }}P\left( k \right):\sum\limits_{k = 1}^n {\sqrt k } \ge \frac{{n\left( {\sqrt n + 1} \right)}}{2} \to {\rm{Adevarat}}\\ \\ \left. {\color{Red} 3} \right)P\left( k \right) \to P\left( {k + 1} \right):\sum\limits_{k = 1}^{n + 1} {\sqrt k } \ge \frac{{\left( {n + 1} \right)\left( {\sqrt {n + 1} + 1} \right)}}{2} \to {\rm{Adevarat}}\\ \\ {\text{Principiul inductiei matematice presupune sa demonstram punctul 3}}.... \end{array}$

$\begin{array}{l} P\left( {k + 1} \right) = \underbrace {\sqrt 1 + \sqrt 2 + ... + \sqrt n }_{P\left( k \right)} + \sqrt {n + 1} \ge \frac{{\left( {n + 1} \right)\left( {\sqrt {n + 1} + 1} \right)}}{2}\\ \\ = \frac{{n\left( {\sqrt n + 1} \right)}}{2} + \sqrt {n + 1} \ge \frac{{\left( {n + 1} \right)\left( {\sqrt {n + 1} + 1} \right)}}{2}\\ \\ \frac{{n\sqrt n + n + 2\sqrt {n + 1} }}{2} \ge \frac{{n\sqrt {n + 1} + n + \sqrt {n + 1} + 1}}{2}\\ \\ n\sqrt n + n + 2\sqrt {n + 1} \ge n\sqrt {n + 1} + n + \sqrt {n + 1} + 1 \Leftrightarrow ... \Leftrightarrow \sqrt n \ge 1 \Leftrightarrow n \ge 1 \to {\rm{adevarat!}}\\ \\ {\text{Deci am demonstrat ca P}}\left( {k + 1} \right) \to {\rm{adevarat!}} \end{array}$

Deci am încheiat !

**Greatmath** Vin Dec 14, 2012 11:47 pm

Sub-punctul a)

Inductie Matematica T754-i10

Soluție oferite de gunty

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