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alex97
Pe: 2013-09-06T12:22:30+03:00 In: In:

Siruri

Buna ziua! Am si eu nevoie de ajutor la 3-4 exercitii dintre acestea.

Multumesc!

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  1. PallMall

        \[\begin{array}{l} 1){({a_n})_{n \ge N}},{a_n} = \frac{{4n}}{{n + 3}} - crescator\\ \to {a_n} \le {a_{n + 1}},\forall n \in N\\ \to \frac{{4n}}{{n + 3}} \le \frac{{4n + 4}}{{n + 4}} \to 4{n^2} + 16n \le 4{n^2} + 16n + 12 \to 0 \le 12\\ 2){({a_n})_{n \ge N}},{a_n} = {n^2} - n - strict\,\,monoton\\ \to {a_n} \le {a_{n + 1}},\forall n \in N\\ \to n(n - 1) \le {(n + 1)^2} - n - 1\\ \to {n^2} - n \le {n^2} + 2n + 1 - n - 1\\ \to 0 \le 2n(n \in {N^*}) \end{array}\]

        \[\begin{array}{l} 4){a_n} = \frac{1}{{\sqrt {n + 1} + \sqrt n }},\forall n \in {N^*}\\ a)dem\,\,{a_n} = \sqrt {n + 1} - \sqrt n \\ {a_n} = \frac{1}{{\sqrt {n + 1} + \sqrt n }} \to {a_n} = \frac{{\sqrt {n + 1} - \sqrt n }}{{(\sqrt {n + 1} + \sqrt n )(\sqrt {n + 1} - \sqrt n )}} \to {a_n} = \sqrt {n + 1} - \sqrt n \end{array}\]

        \[\begin{array}{l} c){a_n} = \sqrt {n + 1} - \sqrt n - strict\,\,descresc\,\,si\,m\arg init\\ {a_n} - strict\,descrescator\\ \to {a_n} > {a_{n + 1}} \to \sqrt {n + 1} - \sqrt n > \sqrt {n + 2} - \sqrt {n + 1} \\ \to 2\sqrt {n + 1} > \sqrt {n + 2} + \sqrt n \\ \to 4n + 4 > n + 2 + n + 2\sqrt {n(n + 2)} \\ \to n + 1 > \sqrt {n(n + 2)} \to {n^2} + 2n + 1 > {n^2} + 2n \to 1 > 0(adv)\\ {a_n} - m\arg init \to \exists n,m \in R/n < {a_n} < m\\ fie\,\,n = 0\,\,si\,\,m = 1\\ \to 0 < \sqrt {n + 1} - \sqrt n < 1(prima\,\,e\,adv)\\ \to \sqrt {n + 1} - \sqrt n < 1\\ \sqrt {n + 1} < 1 + \sqrt n \to n + 1 < 1 + n + 2\sqrt n \to 0 < 2\sqrt n (adv) \end{array}\]

  2. Green eyes
    maestru (V)

    Salut,

    Problema 5:

    Dacă \bl s_n=\frac{n}{n+1} atunci \bl s_{n+1}=\frac{n+1}{n+2}. Pe de altă parte:

    \bl s_{n+1}=a_1+a_2+...+a_n+a_{n+1}=s_n+a_{n+1}. Deci:

    \bl a_{n+1}=s_{n+1}-s_n=\frac{n+1}{n+2}-\frac{n}{n+1}=\frac{(n+1)^2-n^{\small 2}-2\cdot n}{(n+2)\cdot (n+1)}=\frac{1}{(n+2)\cdot (n+1)}. Deci:

    \bl a_{n+1}=\frac{1}{(n+2)\cdot (n+1)}. Înlocuim pe n + 1 cu n:

    \bl a_n=\frac{1}{n\cdot (n+1)}

    Green eyes.

    P.S. Te rog să scrii ceva încercări de a rezolva celelalte probleme, aceasta este una din regulile forumului. Mulţumesc.

  4. gunty
    maestru (V)

        \[<br/> \begin{array}{l}<br/> 1.\;Calculeaza\;S_1 = a_{n + 1} - a_n .\;Daca\;S_1 \ge 0,\;\forall n \in N,\;inseamna\;ca\;\left( {a_n } \right)_{n \ge 0} \;este\;crescator. \\ <br/> 2.\;Calculeaza\;S_2 = a_{n + 1} - a_n .\;Daca\;S_2 > 0,\forall n \in N^* ,\;inseamna\;ca\;\left( {a_n } \right)_{n \ge 1} \;este\;strict\;crescator\;\left( {monoton} \right). \\ <br/> 3.\;a.)\;Sa\;aratam\;ca\;E\left( x \right) > 0,\forall x \ge 4:\;E\left( x \right) = x^2 - 4x + 3 > 0 \Leftrightarrow \left( {x - 2} \right)^2 - 1 > 0 \Leftrightarrow \left( {x - 3} \right)\left( {x - 1} \right) > 0_{\left( {adev.} \right)} \\ <br/> Deoarece\;E\left( x \right) > 0,\forall x \ge 4 \Rightarrow \;a_{n + 1} - a_n = \frac{1}{{E\left( {n + 1} \right)}} > 0,\forall n \ge 4,\;adica\;sirul\;\left( {a_n } \right)_{n \ge 4} \;este\;strict\;crescator. \\ <br/> c.)\;a_n = \sum\limits_{k = 4}^n {\frac{1}{{k^2 - 4k + 3}}} = \sum\limits_{k = 4}^n {\frac{1}{{\left( {k - 3} \right)\left( {k - 1} \right)}}} = \frac{1}{2}\sum\limits_{k = 4}^n {\frac{1}{{k - 3}} - \frac{1}{{k - 1}}} = \\ <br/> = \frac{1}{2}\left( {\left( {\frac{1}{1} - \frac{1}{3}} \right) + \left( {\frac{1}{2} - \frac{1}{4}} \right) + \left( {\frac{1}{3} - \frac{1}{5}} \right) + ... + \left( {\frac{1}{{n - 5}} - \frac{1}{{n - 3}}} \right) + \left( {\frac{1}{{n - 4}} - \frac{1}{{n - 2}}} \right) + \left( {\frac{1}{{n - 3}} - \frac{1}{{n - 1}}} \right)} \right) = \frac{1}{2}\left( {1 + \frac{1}{2} - \frac{1}{{n - 2}} - \frac{1}{{n - 1}}} \right) = \\ <br/> = \frac{1}{2}\left( {\frac{{3\left( {n - 1} \right)\left( {n - 2} \right)}}{{2\left( {n - 1} \right)\left( {n - 2} \right)}} - \frac{{4n - 6}}{{2\left( {n - 1} \right)\left( {n - 2} \right)}}} \right) = \frac{{3n^2 - 13n + 12}}{{4\left( {n - 1} \right)\left( {n - 2} \right)}} = \frac{{\left( {n - 3} \right)\left( {3n - 4} \right)}}{{4\left( {n - 1} \right)\left( {n - 2} \right)}} \\ <br/> \\ <br/> b.)\;Evident,\;ca\;sirul\;este\;marginit\;inferior,\;deoarece\;este\;strict\;crescator\;\left( {deci\;\inf a_n = a_4 } \right).\;Sa\;aratam,\;ca\;a_n < 1,\forall n \ge 4: \\ <br/> \frac{{\left( {n - 3} \right)\left( {3n - 4} \right)}}{{4\left( {n - 1} \right)\left( {n - 2} \right)}} \le 1 \Leftrightarrow 3n^2 - 13n + 12 \le 4n^2 - 12n + 8 \Leftrightarrow n^2 + n - 4 \ge 0 \Leftrightarrow n\left( {n + 1} \right) \ge 4_{\left( {adev.} \right)} ,\;ceea\;ce\;inseamna\;ca\;sirul\;este\;marginit\;superior. \\ <br/> In\;concluzie,\;sirul\;este\;marginit.<E>:)</E> \\ <br/> \end{array}<br/> \]

    Sunt de acord cu Green eyes…vrem sa vedem ca ai incercat si tu. In plus, o alta regula e sa nu postezi mai mult de o problema intr-un subiect.

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