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Thread: inequalities

  1. #1
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    inequalities

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

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

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

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

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

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


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

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

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


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

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

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


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


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

    Check

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

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

    . . We're golden!

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