Home/ Intrebari/Q 78710
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

xor_NTG
maestru (V)
Pe: 2012-11-02T09:44:59+02:00 In: In:

Suma

Cum pot calcula urmatoarea suma?

    \[ \sum\limits_{k = 1}^n {\frac{1}{{k\left( {k + 2} \right)}}} \]

Am incercat niste desompuneri ceva dar nu merge.

Multumesc.

Similare

2 raspunsuri

  1. ali
    maestru (V)

    Este o suma simpla (oricum pana în anul I orice suma o sa ti se pare greoaie)…
    \begin{array}{l} S = \sum\limits_{k = 1}^n {\frac{1}{{k\left( {k + 2} \right)}}} \\ \frac{1}{{k\left( {k + 2} \right)}} = \frac{1}{2}\left( {\frac{1}{k} - \frac{1}{{k + 2}}} \right)\\ k = 1 \Rightarrow \frac{1}{{1 \times 3}} = \frac{1}{2}\left( {\frac{1}{1} - \frac{1}{3}} \right)\\ k = 2 \Rightarrow \frac{1}{{2 \times 4}} = \frac{1}{2}\left( {\frac{1}{2} - \frac{1}{4}} \right)\\ k = 3 \Rightarrow \frac{1}{{3 \times 5}} = \frac{1}{2}\left( {\frac{1}{3} - \frac{1}{5}} \right)\\ ..........\\ k = n \Rightarrow \frac{1}{{n \times \left( {n + 2} \right)}} = \frac{1}{2}\left( {\frac{1}{n} - \frac{1}{{n + 2}}} \right)\\ S = \frac{1}{2}\left( {1 + \frac{1}{2} - \frac{1}{{n + 2}} - \frac{1}{{n + 1}}} \right) = \frac{1}{2} \times \frac{{n\left( {3n + 5} \right)}}{{2\left( {n + 1} \right)\left( {n + 2} \right)}} \end{array}

  2. xor_NTG
    maestru (V)

    Multumesc ali.
    Asta cu calculul sumelor…tine mai mult de experienta (un fel de dexteritate) care se acumuleaza in timp. Eu asa cred.

Raspunde

Raspunde