mateforum.ro Forum Index mateforum.ro

 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

Inegaliate clasa V

 
Post new topic   Reply to topic    mateforum.ro Forum Index -> Clasa a V-a
View previous topic :: View next topic  
Author Message
bogdan2010



Joined: 01 Mar 2011
Posts: 2

PostPosted: Tue Mar 01, 2011 2:25 pm    Post subject: Inegaliate clasa V Reply with quote

Demonstrati ca :
\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\ldots \frac{1}{11}<5
Back to top
View user's profile Send private message
Virgil Nicula
Euler


Joined: 28 Sep 2007
Posts: 672

PostPosted: Tue Mar 01, 2011 4:40 pm    Post subject: Re: Inegaliate clasa V Reply with quote

Quote:
Demonstrati ca \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\ \ldots\ +\frac{1}{11}\ <\ 4


Metoda 1.1. \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\ \ldots\ +\frac{1}{10}+\frac {1}{11}\ <\ \frac 12+\frac 13+8\cdot\frac 14\ =\ 2+\frac 56\ <\ 3\ .

Metoda 1.2. \left\|\begin{array}{ccccc}
\frac 12+\frac 13+\frac 14+\frac 15 & < & 4\cdot\frac 12 & = & 2\\\\\\\\
\frac 16+\frac 17+\frac 18+\frac 19 & < & 4\cdot\frac 16 & = & \frac 23\\\\\\
\frac {1}{10}+\frac {1}{11} & < & 2\cdot\frac {1}{10} & = & \frac 15\end{array}\right\|\ \bigoplus\ \Longrightarrow\ \sum_{k=2}^{11} \frac 1k\ <\ 2+\frac 23+\frac 15\ <\ 3\ .

Metoda 2.1. \frac 12+\frac 13+\frac 14+\ \ldots\ \ \ldots\ \frac {1}{10}+\frac {1}{11}\ <\ \frac 12+9\cdot\frac 13\ =\ 3+\frac 12\ <\ 4\ .

Metoda 2.2. \left\|\begin{array}{ccccc}
\frac 12+\frac 13+\frac 14+\frac 15+\frac 16 & < & 5\cdot\frac 12 & = & \frac 52\\\\\\\\
\frac 17+\frac 18+\frac 19+\frac {1}{10}+\frac {1}{11} & < & 5\cdot\frac 17 & = & \frac 57\end{array}\right\|\ \bigoplus\ \Longrightarrow\ \sum_{k=2}^{11} \frac 1k\ <\ 5\cdot\left(\frac 12+\frac 17\right)=5\cdot\frac {9}{14}\ <\ 4\ .
Back to top
View user's profile Send private message
bogdan2010



Joined: 01 Mar 2011
Posts: 2

PostPosted: Thu Mar 03, 2011 2:49 pm    Post subject: Reply with quote

Folosind ac rationament de ce nu direct :

1/2 + 1/3 +...1/11 < 10*1/2 adica S< 5
Quote:
Back to top
View user's profile Send private message
seby97
Euclid


Joined: 04 Aug 2011
Posts: 31

PostPosted: Sat Oct 29, 2011 1:34 pm    Post subject: Reply with quote

Domnul Nicula a dorit sa arate ca acea suma este chiar mai mica decat 5,oferind ideei bune pentru diverse alte probleme in care nu se va mai aplica metoda directa(adica primul termen x nr de termeni).[/tex]
Back to top
View user's profile Send private message
Integrator



Joined: 01 Nov 2011
Posts: 1

PostPosted: Tue Nov 01, 2011 8:24 am    Post subject: Re: Inegaliate clasa V Reply with quote

bogdan2010 wrote:
Demonstrati ca :
\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\ldots \frac{1}{11}<5

Se observa ca:
\frac{1}{2} +\frac{1}{11}=\frac{13}{22}<\frac{13}{21}

\frac{1}{3} +\frac{1}{10}=\frac{13}{30}<\frac{15}{30}=\frac{1}{2}

\frac{1}{4} +\frac{1}{9}=\frac{13}{36}<\frac{15}{30}=\frac{1}{2}

\frac{1}{5} +\frac{1}{8}=\frac{13}{40}<\frac{13}{39}=\frac{1}{3}

\frac{1}{6} +\frac{1}{7}=\frac{13}{39}<\frac{13}{39}=\frac{1}{3}

Adunand aceste inegalitati rezulta:

\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\ldots \frac{1}{11}<\frac{13}{21}+1+\frac{2}{3}=\frac{48}{21}=\frac{16}{7}.
Observam ca rezulta o inegalitate mult mai buna decat cea din problema caci \frac{16}{7}<5.
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    mateforum.ro Forum Index -> Clasa a V-a All times are GMT + 2 Hours
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum



Powered by phpBB © 2001, 2005 phpBB Group