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dfani511
Pe: 2009-01-11T13:17:58+02:00 In: In:

pls help verificare inductie inegalitate

Pls help si la exercitiul acesta:SA SE verifice prin inductie matematica urmatoarea inegalitate:
2^n>=n^2-1,n>=2
^=ridicare la putere
„>=” =mai mare sau egal

si ar mai fi unu care iar nu il inteleg…1+2+3+…+2<[(n+1)^2]/2

pls si ceva pasi ca nu am inteles din clasa aproape nimic la inductie,si is cam intre nori:((si am lucrare peste cateva zile:(((

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5 raspunsuri

  1. dfani511

    pls help si la me:(

  2. ali
    maestru (V)

    dfani511 wrote: Pls help si la exercitiul acesta:SA SE verifice prin inductie matematica urmatoarea inegalitate:
    2^n>=n^2-1,n>=2

        \[ \begin{array}{l} 2^n \ge n^2 - 1 \\ 1)verificare{\rm sper ca stii s - o faci}{\rm .} \\ {\rm 2)presupunem p(k) adevarat}{\rm .} \\ {\rm 3)p(k)} \to {\rm p(k + 1)}...... \\ {\rm 2}^{{\rm k + 1}} \ge (k + 1)^2 - 1 = k^2 + 2k \\ 2({\rm 2}^{\rm k} ) \ge k^2 - 1 + 2k + 1 \Rightarrow {\rm 2}^{\rm k} + \underbrace {{\rm 2}^{\rm k} \ge k^2 - 1}_{Cer{\mathop{\rm int}} a} + 2k + 1 \Rightarrow {\rm 2}^{\rm k} \ge 2k + 1 \Rightarrow \\ k^2 - 1 \ge 2k + 1 \Rightarrow k^2 - 2 - 2k \ge 0 \\ ia{\rm rez tu inecatia asta,si ai grije cand selectezi valorile pt k > 0}{\rm .} \\ {\rm daca erai in a 11 ti - o rez in 2 randuri,sper ca ai inteles ceva}{\rm .!} \\ \end{array} \]

  4. ali
    maestru (V)

    Pt problema 2 avem:

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 1 + 2 + 3 + ... + n = \frac{{n(n + 1)}}{2} \\
	 {\rm  deci practic tot ce ai de rez este sa dem ca]
	
	

*** Error message:
\begin{array} on input line 10 ended by \end{document}.
leading text: \end{document}
Improper \prevdepth.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
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leading text: \end{document}
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leading text: \end{document}
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leading text: \end{document}
\begin{array} on input line 10 ended by \end{document}.
leading text: \end{document}
Missing } inserted.
leading text: \end{document}
Emergency stop.
  5. dfani511

    Fibonacci wrote: [quote=dfani511]Pls help si la exercitiul acesta:SA SE verifice prin inductie matematica urmatoarea inegalitate:
    2^n>=n^2-1,n>=2

        \[ \begin{array}{l} 2^n \ge n^2 - 1 \\ 1)verificare{\rm sper ca stii s - o faci}{\rm .} \\ {\rm 2)presupunem p(k) adevarat}{\rm .} \\ {\rm 3)p(k)} \to {\rm p(k + 1)}...... \\ {\rm 2}^{{\rm k + 1}} \ge (k + 1)^2 - 1 = k^2 + 2k \\ 2({\rm 2}^{\rm k} ) \ge k^2 - 1 + 2k + 1 \Rightarrow {\rm 2}^{\rm k} + \underbrace {{\rm 2}^{\rm k} \ge k^2 - 1}_{Cer{\mathop{\rm int}} a} + 2k + 1 \Rightarrow {\rm 2}^{\rm k} \ge 2k + 1 \Rightarrow \\ k^2 - 1 \ge 2k + 1 \Rightarrow k^2 - 2 - 2k \ge 0 \\ ia{\rm rez tu inecatia asta,si ai grije cand selectezi valorile pt k > 0}{\rm .} \\ {\rm daca erai in a 11 ti - o rez in 2 randuri,sper ca ai inteles ceva}{\rm .!} \\ \end{array} \]

    miai putea spune de ce ai pus 2(2^k)? si 2^k+2^k nus sigur?,aici nu ma inteles:D pls:D
    ms pt ajutor!

  6. ali
    maestru (V)

    Pai:2^k+1=2*2^k=2^k+2^k,ex:
    2^3=2*2^2=2^2+2^2.
    Sper ca ai intele treaba asta cu puterile//

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