[Algebra]
Problem: Let $n$ be a positive integer such that $\frac{3+4+\cdots+3 n}{5+6+\cdots+5 n}=\frac{4}{11}$. Compute $\frac{2+3+\cdots+2 n}{4+5+\cdots+4 n}$.

Solution: In simplifying the numerator and denominator of the left side of the equation, notice that

$$
\begin{aligned}
k+(k+1)+\cdots+k n & =\frac{1}{2}(k n(k n+1)-k(k-1)) \\
& =\frac{1}{2}(k(n+1)(k n-k+1))
\end{aligned}
$$

This identity allows the given equation to be transformed:

$$
\begin{aligned}
\frac{3(n+1)(3 n-3+1)}{5(n+1)(5 n-5+1)} & =\frac{4}{11} \\
\frac{3(n+1)(3 n-2)}{5(n+1)(5 n-4)} & =\frac{4}{11} \\
\frac{3 n-2}{5 n-4} & =\frac{20}{33}
\end{aligned}
$$

Solving this last equation yields $n=14$. Using the same identity twice more, for $n=14$ and $k=2$ and $k=4$, the desired quantity is $\frac{2(2 n-1)}{4(4 n-3)}=\frac{\mathbf{2 7}}{\mathbf{1 0 6}}$.

Final answer: $\frac{27}{106}$