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Pe: 2010-08-11T18:56:45+03:00 In: In:

suma

sa se calculeze suma

    \[\sum\limits_{k = 1}^n {\frac{6}{{k\left( {k + 1} \right)\left( {k + 3} \right)}}} \]

eu am incercat ceva de genul

    \[\frac{6}{{k\left( {k + 1} \right)\left( {k + 3} \right)}} = 2\left( {\frac{1}{{k(k + 1)}} - \frac{1}{{(k + 1)(k + 3)}}} \right)\]

    \[{S_n} = 2\left( {\frac{1}{{1*2}} - \frac{1}{{2*4}}} \right) + 2\left( {\frac{1}{{2*3}} - \frac{1}{{2*4}}} \right) + ... + 2\left( {\frac{1}{{k(k + 1)}} - \frac{1}{{(k + 1)(k + 3)}}} \right) = \]

dar nu iasa.., poate nui buna metoda prin care am rezolvat?

Similare

3 raspunsuri

  1. maestru (V)
    Indicatie

        $$ \frac{1} {{k\left( {k + 1} \right)\left( {k + 3} \right)}} = \frac{{\text{1}}} {{{\text{3k}}}} - \frac{{\text{1}}} {{{\text{2}}\left( {{\text{k}} + {\text{1}}} \right)}} + \frac{{\text{1}}} {{{\text{6}}\left( {{\text{k}} + {\text{3}}} \right)}} $$

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  2. user (0)

    Fibonacci multumesc, dar tot nam reusit sa ma prind cum vine mai departe 🙁

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  4. maestru (V)

    Totusi, cred ca merge mai repede cu forma ta:

        \[ \begin{array}{l} 2\sum {\left[ {\frac{1}{{k\left( {k + 1} \right)}} - \frac{1}{{\left( {k + 1} \right)\left( {k + 3} \right)}}} \right]} = 2\sum {\frac{1}{{k\left( {k + 1} \right)}}} - 2\sum {\frac{1}{{\left( {k + 1} \right)\left( {k + 3} \right)}}} \\ \sum {\frac{1}{{k\left( {k + 1} \right)}}} = \sum {\left( {\frac{1}{k} - \frac{1}{{k + 1}}} \right)} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1 - \frac{1}{{n + 1}} \\ \end{array} \]

    La fel se procedeaza si pt cealalta suma.
    Nu am verificat forma ta, daca este determinata corect atunci urmeaza pasi indicati de mine.

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