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Math Help - inequalities

  1. #1
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    inequalities

    \sum^{n}_{r=1}r^3>18000

    \frac{1}{4}n^2(n+1)^2>18000

    i am not sure how to solve this . Thanks for helping .
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  2. #2
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    Hello, thereddevils!

    \sum^{n}_{r=1}r^3\:>\:18,\!000

    \frac{1}{4}n^2(n+1)^2\:>\:18,\!000

    i am not sure how to solve this . Thanks for helping .
    Note that n is a positive integer.
    . . Hence, n(n+1) is also positive.


    Multiply by 4: . n^2(n+1)^2 \:>\:72,\!000

    Take the square root: . n(n+1) \:>\:\sqrt{72,\!000} \:=\:120\sqrt{5}

    And we have: . n^2 + n - 120\sqrt{5}\:>\:0


    We have: . y \:=\:x^2 + x - 120\sqrt{5} . . . an up-opening parabola.
    . . It is positive on the "outside" of its x-intercepts.

    So we solve: . x^2 + x - 120\sqrt{5} \:=\:0
    . . Quadratic Formula: . x \;=\; \frac{-1 \pm\sqrt{1 + 480\sqrt{5}}}{2} \;\approx\;\begin{Bmatrix}15.89 \\ \text{-}16.89\end{Bmatrix}

    Hence, y is positive for: . (x \:\leq \:-16.89) \;\cup \;(x \:\geq \:15.89)


    . . Therefore: . \boxed{n \:\geq\: 16}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Check

    n = 15\!:\;\;\sum^{15}_{r=1}r^3 \:=\:14,\!400

    n = 16\!:\;\;\sum^{16}_{r-1}r^3 \:=\:18,\!496

    . . We're golden!

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