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mihaela_0205
Pe: 2012-11-06T13:46:21+02:00 In: In:

Algebra!!!

Calculati :
1/5 + 7/10 + 8/15 + … + 50/225 – (1/2 + 1/3 + 1/4 + … + 1/45)

Prima oara m-am gandit sa elimin paranteza si s-ar fi redus 1/5 cu -1/5 ..
sau as fi facut suma din paranteza..

Dar acum am incercat sa dau factor comun pe 1/5 , si ar veni :
1/5(7/2 + 8/3+ 9/4 +…+50/45) – (1/2+1/3+1/4+…+1/45)
observ ca termenii din ambele paranteze au acelasi numitor, dar ce pot sa fac mai departe?

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2 raspunsuri

  1. ali
    maestru (V)

    \begin{array}{l} A = \frac{1}{5} + \frac{7}{{10}} + \frac{8}{{15}} + \cdots + \frac{{50}}{{225}} - \left( {\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{{45}}} \right) = \frac{1}{5}\left( {1 + \frac{7}{2} + \frac{8}{3} + \cdots + \frac{{50}}{{45}}} \right) - \frac{1}{5}\left( {\frac{5}{2} + \frac{5}{3} + \cdots + \frac{5}{{45}}} \right)\\ = \frac{1}{5}\left( {1 + \frac{7}{2} - \frac{5}{2} + \frac{8}{3} - \frac{5}{3} + \frac{9}{4} - \frac{5}{4} + \cdots + \frac{{50}}{{45}} - \frac{{45}}{{45}}} \right) = \frac{1}{5}\left( {1 + \underbrace {1 + 1 + \cdots + 1}_{44}} \right) = 9 \end{array}
    Cum ti se pare? 😀

  2. mihaela_0205

    ali wrote: \begin{array}{l} A = \frac{1}{5} + \frac{7}{{10}} + \frac{8}{{15}} + \cdots + \frac{{50}}{{225}} - \left( {\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{{45}}} \right) = \frac{1}{5}\left( {1 + \frac{7}{2} + \frac{8}{3} + \cdots + \frac{{50}}{{45}}} \right) - \frac{1}{5}\left( {\frac{5}{2} + \frac{5}{3} + \cdots + \frac{5}{{45}}} \right)\\ = \frac{1}{5}\left( {1 + \frac{7}{2} - \frac{5}{2} + \frac{8}{3} - \frac{5}{3} + \frac{9}{4} - \frac{5}{4} + \cdots + \frac{{50}}{{45}} - \frac{{45}}{{45}}} \right) = \frac{1}{5}\left( {1 + \underbrace {1 + 1 + \cdots + 1}_{44}} \right) = 9 \end{array}
    Cum ti se pare? 😀

    Multumesc foarte mult! Nici nu stiu cum as putea sa va multumesc pentru toate raspunsurile date pana acum 😀

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