25~26高三下·江苏扬州一模·第14题

$\dfrac{2n}{n+1}$

由函数$f(x) = {x[x]}$在定义域$[0,n)\ (n \in \mathbb{N}^*)$上的值域为$C_n$，记$C_n$中元素的个数为$a_n$.

当$n=1$时，$x \in [0,1)$，可得$[x] = 0$，$x[x] = 0$，${x[x]} = 0$，即$a_1 = 1$；

当$n=2$时，$x \in [0,2)$，可得$[x] = 0$或$1$，$x[x] = 0$或$x$，${x[x]} = 0$或$1$或$2$，即$a_2 = 3$；

当$n=3$时，$x \in [0,3)$，可得$[x] = 0$或$1$或$2$，$x[x] = 0$或$x$或$2x$，${x[x]} = 0$或$1$或$2$或$3$或$4$或$5$或$6$，即$a_3 = 6$.

当$n = k-1\ (k \in \mathbb{N}^, k \ge 2)$时，函数$f(x) = {x[x]}$在定义域$[0,k-1)\ (k \in \mathbb{N}^, k \ge 2)$上的值域为$C_{k-1}$，记$C_{k-1}$中元素的个数为$a_{k-1}$；

当$n = k$时，函数$f(x) = {x[x]}$在定义域$[0,k)\ (k \in \mathbb{N}^*)$上的值域为$C_k$，记$C_k$中元素的个数为$a_k$.

设$x \in [k-1,k)$，则$[x] = k-1$，$(k-1)^2 \le x[x] = (k-1)x < k(k-1)$，

所以$a_k = a_{k-1} + k(k-1) – (k-1)^2 + 1$$= a_{k-1} + k\ (k \in \mathbb{N}^*, k \ge 2)$，

则可得递推关系：$a_n = a_{n-1} + n\ (n \in \mathbb{N}^*, n \ge 2)$.

所以$a_n = a_1 + (a_2 – a_1) + (a_3 – a_2) + \dots + (a_n – a_{n-1})$$= 1 + 2 + 3 + \dots + n = \dfrac{n(n+1)}{2}\ (n \in \mathbb{N}^*, n \ge 2)$，

当$n=1$时，$a_1 = \dfrac{1 \times 2}{2} = 1$成立，则$a_n = \dfrac{n(n+1)}{2}\ (n \in \mathbb{N}^*)$.

则$\dfrac{1}{a_n} = \dfrac{2}{n(n+1)} = 2\left(\dfrac{1}{n} – \dfrac{1}{n+1}\right)$，

所以$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dots + \dfrac{1}{a_n}$$= 2\left(1 – \dfrac{1}{2} + \dfrac{1}{2} – \dfrac{1}{3} + \dots + \dfrac{1}{n} – \dfrac{1}{n+1}\right)$$= 2\left(1 – \dfrac{1}{n+1}\right) = \dfrac{2n}{n+1}$.