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m_radw.ro
Pe: 2013-01-16T19:18:12+02:00 In: In:

Mate:((

Sa se demonstreze prin inductia matematicii :
4^n/n+1<(2n)!/(n!)^2

Nu am reusit sa inteleg deloc inductia asta 😯
Va multumesc anticipat pt raspuns!!!

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  1. ali
    maestru (V)

    Inegalitatea nu este adevarat pentru n=1… deci…. cred ca ai uitat sa specifici acest lucru (pentru tine probabil este o observatie banala dar d.p.d.v.d matematic da totul peste cap).
    Rezolvarea absolut completa (si nemeritata) este:
    \begin{array}{l} {p_k}:\frac{{{4^k}}}{{k + 1}} < \frac{{\left( {2k} \right)!}}{{{{\left( {k!} \right)}^2}}}\\ \bullet \,{\rm{verificare: }}\left\{ \begin{array}{l} {p_1}:\frac{{{4^1}}}{{1 + 1}} = \frac{4}{2} = 2 < \frac{{\left( {2 \times 1} \right)!}}{{{{\left( {1!} \right)}^2}}} = \frac{{2!}}{{1!}} = 2 \to 2 < 2 \to {\rm{fals}}\\ {p_2}:\frac{{{4^2}}}{{2 + 1}} = \frac{{16}}{3} = 5.3 < \frac{{\left( {2 \times 2} \right)!}}{{{{\left( {2!} \right)}^2}}} = \frac{{4!}}{{2! \times 2!}} = \frac{{24}}{4} = 6 \to 5.3 < 6 \to {\rm{adevarat}}\\ {p_3}:\frac{{{4^3}}}{{3 + 1}} = \frac{{64}}{4} = 16 < \frac{{\left( {2 \times 3} \right)!}}{{{{\left( {3!} \right)}^2}}} = \frac{{6!}}{{3! \times 3!}} = \frac{{720}}{{36}} = 20 \to 16 < 20 \to {\rm{adevarat}} \end{array} \right.\\ \bullet \,\,{\rm{prespunem }}{p_k}:\frac{{{4^k}}}{{k + 1}} < \frac{{\left( {2k} \right)!}}{{{{\left( {k!} \right)}^2}}} \to {\rm{adevarat pentru k numere}}{\rm{.}}\\ \bullet \,\,{\rm{demonstram (daca este sau nu e adevarat implicatia) }}{p_k} \to {p_{k + 1}} \to {\rm{adevarat}}.\\ {p_{k + 1}}:\frac{{{4^{k + 1}}}}{{k + 1 + 1}} = \frac{{{4^{k + 1}}}}{{k + 2}} < \frac{{\left( {2\left( {k + 1} \right)} \right)!}}{{{{\left( {\left( {k + 1} \right)!} \right)}^2}}} = \frac{{\left( {2k + 2} \right)!}}{{{{\left( {\left( {k + 1} \right)!} \right)}^2}}}\\ {\rm{Demonstratie:}}\\ \frac{{{4^{k + 1}}}}{{k + 2}} = \frac{{4\left( {k + 1} \right)}}{{k + 2}}{ \times \left( {\frac{{{4^k}}}{{k + 1}}} \right)}\limits_{ = {p_k}} < \frac{{4\left( {k + 1} \right)}}{{k + 2}} \times \frac{{\left( {2k} \right)!}}{{{{\left( {k!} \right)}^2}}}\,\,< \limits^{{\rm{demonstram}}} \,\,\frac{{\left( {2k + 2} \right)!}}{{{{\left( {\left( {k + 1} \right)!} \right)}^2}}} = \frac{{\left( {2k} \right)!\left( {2k + 1} \right)\left( {2k + 2} \right)}}{{{{\left( {k!} \right)}^2}{{\left( {k + 1} \right)}^2}}}\\ 4{\left( {k + 1} \right)^2} < 2\left( {2k + 1} \right)\left( {k + 2} \right) \Leftrightarrow 2k > 0 \to true\\ {\rm{Deci am demonstrat pasul k + 1 folosindu - ne de pasul k }}...{\rm{ceea ce}}\\ {\rm{incheie si demonstratia noastra}}{\rm{.}} \end{array}

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