# inequalities

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

Quote:

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

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

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

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Check

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

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

. . We're golden!