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steve jobs

17/03/2017 at 14:12
Answers
3
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Evaluate \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}.\)


rational numbers


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  • ...
    Trần Quang Linh 17/03/2017 at 15:59

    A=\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

    128A=64+32+16+8+4+2+1

    128A=127

    A=\(\dfrac{127}{128}\)

    steve jobs selected this answer.
  • ...
    Love people Name Jiang 17/03/2017 at 18:43

    For : \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)

    \(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+.....+\dfrac{1}{128}\)

    \(\Rightarrow2A-A=1-\dfrac{1}{128}\)

    \(\Rightarrow A=\dfrac{127}{128}\)

  • ...
    FA KAKALOTS 03/02/2018 at 12:46

    For : A=12+14+18+132+164+1128

    ⇒2A=1+12+14+.....+1128

    ⇒2A−A=1−1128

    ⇒A=127128


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