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    please make solutions in white --->like this<---
    if you can do this it feels very satisfying because there's a very subtle solution
    ant try not to cheat by looking at others!

    (I hope I've typed it out correctly)
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    Like this you mean? (Or do you want it simplified further or solved?)
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    dvs: shouldn't that be (n^2 + n)^(1/3) rather than (n^2 + 2n)^(1/3) ?
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    Yes, it should. I mistyped it!
    • Thread Starter
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    I believe it can be simplifies further if I have types it out correctly.

    hint:
    factorise (a^3-b^3)
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    Using x^3 - y^3 = (x - y)(x^2 + xy + y^2),

    1
    = (n + 1) - n
    = cbrt(n + 1)^3 - cbrt(n)^3
    = [cbrt(n + 1) - cbrt(n)] [cbrt(n + 1)^2 + cbrt(n + 1)cbrt(n) + cbrt(n)^2]
    = [cbrt(n + 1) - cbrt(n)] [cbrt(n^2 + 2n + 1) + cbrt(n^2 + n) + cbrt(n^2)]

    (sum from n = 1 to 100) 1 / [cbrt(n^2 + 2n + 1) + cbrt(n^2 + n) + cbrt(n^2)]
    = (sum from n = 1 to 100) [cbrt(n + 1) - cbrt(n)]
    = cbrt(101) - 1
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    yeah that's right. another method:

    a^2 + ab + b^2 = (a^3 - b^3)/(a - b)
    1/(a^2 + ab + b^2) = (a - b)/(a^3 - b^3)

    so 1/[3rt{1} + 3rt{2} + 3rt{4}] = (3rt 2 - 3rt 1)
    ect. onto 1/[3rt{10000} + 3rt{10100} + 3rt{10201}] = (3rt 101 - 3rt 100)

    all others in between cancel to leave 3rt 101 - 1
 
 
 
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Updated: February 3, 2005

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