1. ## Quickie #4

$\text{Given: }\;\frac{1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3}{2^3 + 4^3 + 6^3 + \cdots + (2n)^3} \;=\;\frac{199}{242}$

. . $\text{find }n.$

2. $\sum_{k=1}^{n}(2k-1)^{3}=2n^{4}-n^{2}$

$\sum_{k=1}^{n}(2k)^{3}=2n^{2}(n+1)^{2}$

$\frac{2n^{4}-n^{2}}{2n^{2}(n+1)^{2}}=\frac{2n^{2}-1}{2(n+1)^{2}}$

$\frac{2n^{2}-1}{2(n+1)^{2}}=\frac{199}{242}$

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$86n^{2}-796n-640=0$

$2(n-10)(43n+32)=0$

$\boxed{n=10} \;\ or \;\ \frac{-32}{43}$

3. Nice solution, Galactus!

By the way, these "Quickie" solutions are not meant to be "better" than yours.
Most of them employ a "trick" of some sort ... some well-known, others obscure.
I found each to be an eye-opening experience.

$\text{Given: }\;\frac{1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3}{2^3 + 4^3 + 6^3 + \cdots + (2n)^3} \;=\;\frac{199}{242}$

. . $\text{find }n.$

We are expected to know: . $\sum^n_{k=1}k^3\:=\:\left[\sum^n_{k=1}k\right]^2\:=\:\frac{n^2(n+1)^2}{4}$

. . and that: .If $\frac{a}{b} = \frac{c}{d}$, then: . $\frac{a+b}{b} \:=\:\frac{c+d}{d}$

So we have: . $\frac{1^3 + 2^3 + 3^3 + \cdots + (2n)^3}{2^3(1^3 + 2^3 + 3^3 + \cdots + n^3)} \;=\;\frac{199+242}{242}$

Then: . $\frac{\frac{(2n)^2(2n+1)^2}{4} }{2^3\!\cdot\!\frac{n^2(n+1)^2}{4}} \:=\:\frac{441}{242} \quad\Rightarrow\quad \frac{4n^2(2n+1)^2}{8n^2(n+1)^2} \:=\:\frac{21^2}{2(11^2)}$

. . Hence: . $\left(\frac{2n+1}{n+1}\right)^2 \:=\:\left(\frac{21}{11}\right)^2$

It concludes with: "Clearly, $n = 10.$
. . The negative square root gives a negative value for $n.$"