# inequalities

• Sep 2nd 2009, 06:27 AM
thereddevils
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 .
• Sep 2nd 2009, 07:33 AM
Soroban
Hello, thereddevils!

Quote:

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

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

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

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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!