10. Let $\alpha_{n}$ be a real root of the cubic equation $n x^{3}+2 x-n=0$, where $n$ is a positive integer. If $\beta_{n}=\left\lfloor(n+1) \alpha_{n}\right\rfloor$ for $n=234 \cdots$, find the value of $\frac{1}{1006} \sum_{k=2}^{2013} \beta_{k}$.
10. Answer: 2015
Let $f(x)=n x^{3}+2 x-n$. It is easy to see that $f$ is a strictly increasing function for $n=234 \cdots$ Further,
$$
f\left(\frac{n}{n+1}\right)=n\left(\frac{n}{n+1}\right)^{3}+2\left(\frac{n}{n+1}\right)-n=\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$. Thus, the only real root of the equation $n x^{3}+2 x-n=0$ for $n \geq 2$ is located in the interval $\left(\frac{n}{n+1} 1\right)$. We must have
$$
\frac{n}{n+1}<\alpha_{n}<1 \Longrightarrow n<(n+1) \alpha_{n}<n+1
$$
so that $\beta_{n}=\left\lfloor(n+1) \alpha_{n}\right\rfloor=n$ for all $n \geq 2$. Consequently,
$$
\frac{1}{1006} \sum_{k=2}^{2013} \beta_{k}=\frac{1}{1006} \sum_{k=2}^{2013} k=\frac{1}{1006} \cdot \frac{2012}{2}(2+2013)=2015
$$