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estudier

  • 2 years ago

f(n) = integer nearest sqrt(n) 1/f(1) + 1/f(2) + ...........+1/f(9999) +1/f(10000) = ?

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  1. mahmit2012
    • 2 years ago
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    |dw:1350239055071:dw|

  2. estudier
    • 2 years ago
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    That looks promising...

  3. estudier
    • 2 years ago
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    So when is f(n) = m?

  4. experimentX
    • 2 years ago
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    *

  5. estudier
    • 2 years ago
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    Get rid of them nasty radicals as well.....

  6. sara12345
    • 2 years ago
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    f(n) would range from 1 to 100

  7. sara12345
    • 2 years ago
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    how to get rid of radical

  8. estudier
    • 2 years ago
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    f(n) = a when (a-1/2)^2 < n < (a+1/2)^2 (I'm using a)

  9. sara12345
    • 2 years ago
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    ok

  10. estudier
    • 2 years ago
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    Think I will leave this one up overnight in case people want to think more about it....

  11. mukushla
    • 2 years ago
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    and im watching :)

  12. sara12345
    • 2 years ago
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    does this involve integrals ?

  13. estudier
    • 2 years ago
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    Certainly not! (I'm not that fond of calculus)

  14. sara12345
    • 2 years ago
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    2a is the window for n until a increments

  15. sara12345
    • 2 years ago
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    is this right direction

  16. estudier
    • 2 years ago
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    Yes! Now u have the pattern, away u go....:-)

  17. sara12345
    • 2 years ago
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    so f(n) = a a = 1 ; 2.1 times a = 2 ; 2.2 times .... a = 99; 2.99 times a = 100; 2.100/2 times

  18. estudier
    • 2 years ago
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    Then the sum is ?

  19. estudier
    • 2 years ago
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    The last bit is just some counting (don't overcount)

  20. sara12345
    • 2 years ago
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    lol its easy

  21. sara12345
    • 2 years ago
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    2a.1/a a would cancel out ! 99*2 + 1 = 199

  22. estudier
    • 2 years ago
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    Yes, well done...

  23. sara12345
    • 2 years ago
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    nice exercise thnk you :)

  24. estudier
    • 2 years ago
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    yw

  25. estudier
    • 2 years ago
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    Since it has been solved I will close it after all....

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