Home/ Intrebari/Q 75301
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

LoGouT
Pe: 2012-01-14T14:11:09+02:00 In: In:

INDUCTIE

Buna,am o mica problema cu o inductie cu dubla inegalitate:

    \[ 1 < \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{1}{{n + 3}} + ....... + \frac{1}{{3n + 1}} < 2,n \in N* \]

Am incercat prin inductie la fiecare in parte dar nu mi-a iesit ,alta metoda exista?Multumesc anticipat
Scuzati dublu-post dar cand am dat trimite mi-a dat Internet Connection cannot display the webpage si am crezut ca nu sa trimis

Similare

4 raspunsuri

  1. PhantomR
    expert (VI)

    Salut!:)

    Voi incerca sa demonstrez prima inegalitate. Verificam mai intai pentru n=1 si obtinem 1<\frac{1}{2}+\frac{1}{3}+\frac{1}{4} |\cdot 12 \Leftrightarrow 12<6+4+3 \Leftrightarrow 12<13, adevarat. Presupunem ca pentru k\leq n avem 1<\frac{1}{k+1}+...+\frac{1}{3k+1} si trebuie sa demonstram ca (aici probabil ati gresit, la trecerea de la k la k+1) 1<\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{3(k+1)+1}=\frac{1}{k+2}+...+\frac{1}{3k+4}=(\frac{1}{k+1}+...+\frac{1}{3k+1})+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}. Dar din ipoteza de inductie avem \frac{1}{k+1}+...+\frac{1}{3k+1}>1 si deci mai ramane de demonstrat ca \frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}>0 \Leftrightarrow \frac{1}{3k+2}+\frac{1}{3k+4}>\frac{1}{k+1}-\frac{1}{3k+3} \Leftrightarrow \frac{1}{3k+2}+\frac{1}{3k+4}>\frac{2}{3k+3} \Leftrightarrow \frac{3k+4+3k+2}{(3k+2)(3k+4)}>\frac{2}{3(k+1)} \Leftrightarrow \frac{6(k+1)}{(3k+2)(3k+4)}>\frac{2}{3(k+1)} \Leftrightarrow 18(k+1)^2>2(3k+2)(3k+4) \Leftrightarrow 9(k+1)^2>(3k+2)(3k+4) \Leftrightarrow 9k^2+18k+9>9k^2+12k+6k+8 \Leftrightarrow 9>8, adevarat. Asadar, conform principiului inductiei matematice inegalitatea este adevarata.

    Procedati analog pentru cea din dreapta si ar trebui sa va dea. Succes!

    Imi cer iertare daca m-am lungit prea mult, dar am vrut sa fie cat mai detaliat.

  2. LoGouT

    Multumesc,gresisem atunci cand am scris P(K+1)

  4. PhantomR
    expert (VI)

    CU cea mai mare placere! O sugestie: la trecerea de la k la k+1 incercati sa il inlocuiti pe k din ipoteza cu k+1.

  5. LoGouT

    O mica problema:

    *** QuickLaTeX cannot compile formula:

	\[
	\begin{array}{l}
	 \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \frac{1}{{n + 3}} + ..... + \frac{1}{{3n + 1}} < 2 \\
	 {\rm Fac P(1) si imi da adevarat} \\
	 {\rm Presupun P(K) A si demonstrez P(K + 1)}{\rm  A} \\
	 P(K)]
	
	
	

*** 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}
Missing } inserted.
leading text: \end{document}
Missing \cr inserted.
leading text: \end{document}
Missing $ inserted.
leading text: \end{document}
You can't use `\end' in internal vertical mode.
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.
Raspunde

Raspunde