回復 2# d3054487667 的帖子
第13題
\(\begin{align}
& {{a}_{n}}=3{{n}^{2}}-3n-1 \\
& {{a}_{3n}}=27{{n}^{2}}-9n-1 \\
& {{a}_{2n}}=12{{n}^{2}}-6n-1 \\
& \underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt[3]{{{a}_{3}}+{{a}_{6}}+{{a}_{9}}+\cdots +{{a}_{3n}}}-\sqrt[3]{{{a}_{2}}+{{a}_{4}}+{{a}_{6}}+\cdots +{{a}_{2n}}}}{n} \\
& =\underset{n\to \infty }{\mathop{\lim }}\,\sqrt[3]{\frac{27\times \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-9\times \frac{n\left( n+1 \right)}{2}-1}{{{n}^{3}}}}-\sqrt[3]{\frac{12\times \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}-6\times \frac{n\left( n+1 \right)}{2}-1}{{{n}^{3}}}} \\
& =\sqrt[3]{9}-\sqrt[3]{4} \\
\end{align}\)