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roberta_12
Pe: 2012-04-12T13:34:35+03:00 In: In:

Help me!

Calculati n=2003+2*(1+2+3+…+2002)

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5 raspunsuri

  1. forumus-matematicus

    Pasul 1) 1+2+3+…+2002 = [2002* (2002+1)]:2= (2002*2003):2=
    = 1001*2003=2005003
    Pasul 2)2*2005003= 4010006
    Pasul 3) 2003 + 4010006 = 4012009
    => n=40120009

  2. Minecraft
    maestru (V)

    raluca8 wrote: \it{n = 2003+2\cdot \frac{2002\cdot2003}{2}\\\;\\Simplificam cu 2.\\\;\\n = 2003+2002\cdot2003 = 2003\cdot(1 + 2002) = 2003\cdot 2003 = 4 012 009.}

        \[ sau...{\rm }2003^2 \]

  4. cata99ionut

    1+2+…+n=n(n+1)/2

  5. forumus-matematicus

    Cu acest prilej de sarbatoare va urez : PASTE FERICIT!!!

  6. cata99ionut

    PASTE FERICIT! 😀

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