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Minecraft
maestru (V)
Pe: 2012-10-27T17:23:40+03:00 In: In:

Sir de numere rationale

    \[\begin{array}{l} {\rm{Daca }}{a_1},{a_2},...,{a_{2012}} \in {Q_ + }{\rm{ si }}\frac{1}{{{a_1} + 1}} + \frac{2}{{{a_2} + 2}} + \frac{3}{{{a_3} + 3}} + .... + \frac{{2012}}{{{a_{2012}} + 2012}} = 2011{\rm{ }}\\ {\rm{Calculati }}\frac{{{a_1}}}{{{a_1} + 1}} + \frac{{{a_2}}}{{{a_2} + 2}} + \frac{{{a_3}}}{{{a_3} + 3}} + .... + \frac{{{a_{2012}}}}{{{a_{2012}} + 2012}} \end{array}\]

Am fost la Dan Barbilian la Calarasi si nu am prea stiut-o.
Ma puteti ajuta sa o verific?

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3 raspunsuri

  1. ali
    maestru (V)

    Sper sa nu fie târziu 🙁
    \begin{array}{l} \frac{{{a_i}}}{{{a_i} + i}} = \frac{{{a_i} + i - i}}{{{a_i} + i}} = 1 - \frac{i}{{{a_i} + i}}\\ \sum {\frac{{{a_i}}}{{{a_i} + i}}} = 2012 - \underbrace {\sum {\frac{i}{{{a_i} + i}}} }_{ = 2011} = 1 \end{array}
    Dupa cum vezi … idea este extrem de simpla 😀

    • 0
  2. edy8
    maestru (V)

    Minecraft wrote:

        \[\begin{array}{l} {\rm{Daca }}{a_1},{a_2},...,{a_{2012}} \in {Q_ + }{\rm{ si }}\frac{1}{{{a_1} + 1}} + \frac{2}{{{a_2} + 2}} + \frac{3}{{{a_3} + 3}} + .... + \frac{{2012}}{{{a_{2012}} + 2012}} = 2011{\rm{ }}\\ {\rm{Calculati }}\frac{{{a_1}}}{{{a_1} + 1}} + \frac{{{a_2}}}{{{a_2} + 2}} + \frac{{{a_3}}}{{{a_3} + 3}} + .... + \frac{{{a_{2012}}}}{{{a_{2012}} + 2012} } \end{array}\]


    \rm{\Large\bl \frac{a_1}{a_1+1} +\frac{a_2}{a_2+2} + ... +\frac{a_{2012}}{a_{2012}+2012}= \frac{a_1+1-1}{a_1+1}+\frac{a_2+2-2}{a_2+2}+ ... +\frac{a_{2012}+2012-2012}{a_{2012}+2012}=\\\;\\= \underbrace{1+1+ ... +1}_{2012 termeni}-(\frac{1}{a_1+1}+\frac{2}{a_2+2}+ ... +\frac{2012}{a_{2012}+2012}) =2012-2011=1.}

    • 0
  4. Minecraft
    maestru (V)

    Multumesc Ali si Edy! Se dovedeste a n-a oara ca nu tot ce pare stufos e si complicat!

    • 0
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