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Urmator
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Pe: 2013-11-30T19:30:52+02:00 In: In:

Algebra

Imi puteti da o rezolvare la exercitiul urmator va rog ?
Calculati :
a) 1/2 + 1/2^2 + 1/2^3 + … + 1/2^40.
b) (3/1 + 5/2 + 7/3 + 9/4 + … + 1999/999) – ( 1+ 1/2 + 1/3 + 1/4 + … + 1/999) .
c) 1/1*2 + 1/2*3 + 1/3*4 + … + 1/1998*1999.
d) (4/3 + 5/4 + 6/5 +7/6) – ( 1/3 + 1/4 +1/5 + 1/6 )

Multumesc !

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2 raspunsuri

  1. tata2008
    maestru (V)

    \rm{\\ a) 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^{40}. \\ b) (3/1 + 5/2 + 7/3 + 9/4 + ... + 1999/999) - ( 1+ 1/2 + 1/3 + 1/4 + ... + 1/999)


    \rm{\bl\\ a). Cauta pe forum ! Vei gasi indrumari pt. aceste tipuri de exercitii ... ;\\ b). Incearca sa calculezi diferentele fractiilor care au acelasi numitor !\\ (3/1-1/1)+(5/2-1/2)+(7/3-1/3)+(9/4-1/4)+ ... + (1999/999-1/999) = \underbrace{ 2+2+2+ ... +2 }_{2 de 999-ori} =

    d) (4/3 + 5/4 + 6/5 +7/6) – ( 1/3 + 1/4 +1/5 + 1/6 )


    \rm{\bl\\ Se procedeaza asemanator ca la punctul b). Succes !

  2. tata2008
    maestru (V)

    c) 1/1*2 + 1/2*3 + 1/3*4 + … + 1/1998*1999.

    \rm{\bl\\ Foloseste descompunerea: \fbox{ \frac{1}{k(k+1} = \frac{1}{k}-\frac{1}{k+1} } !\\ Vei obtine: \frac{1}{1} - \underbrace{\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+ ... +\frac{1}{1998}}_{=0}-\frac{1}{1999}= ...

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