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maria_dobrila
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Pe: 2012-10-29T18:59:06+02:00 In: In:

algebra

2 3 4 5 2010 2011 2012
a=2011(-1)+2010(-1)+2009(-1)+2008(-1)+…+3(-1) + 2(-1) + 1(-1)
–––––––––––––––––––––––
1 1 1 1 1
– + ––+ ––+ –––- +––––-
1*2 2*3 3*4 2010*2011 2011+2012

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  1. dan
    expert (VI)

    maria_dobrila wrote: 2 3 4 5 2010 2011 2012
    a=2011(-1)+2010(-1)+2009(-1)+2008(-1)+…+3(-1) + 2(-1) + 1(-1)
    –––––––––––––––––––––––
    1 1 1 1 1
    – + ––+ ––+ –––- +––––-
    1*2 2*3 3*4 2010*2011 2011*2012

    Va multumesc pentru ajutor.


    Incearca sa descrii prin cuvinte „enuntul problemei/exercitiului”. La primul exercitiu nu se intelege nimic iar la a 2-a se pare ca este vorba de: \rm{S=\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+...+\frac{1}{2011*2012} = ?\\ La exercitiul 2 se foloseste \fbox{ \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1)}}\\ adica: \| \frac{1}{1*2}=1/1-1/2\\ \frac{1}{2*3}=1/2-1/3\\ ......................................\\ \frac{1}{2011*2012}=1/2011-1/2012\\ suma lor va fi: ... \fbox{S = 1/1-1/2012 = 2011/2012\surd\surd}\\ ..........................................\\ ..........................................\\\bl

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