1069

设数列 ${\displaystyle \{a_n\}}$ 满足 ${\displaystyle a_1=1,a_{n+1}={\displaystyle a_n+\frac{1}{a_n}-\frac{1}{a_n+1}(n\ge 1)}}$, 则 ${\displaystyle [a_{100}]=}$ _____. (其中 ${\displaystyle [x]}$ 表示不超过实数 ${\displaystyle x}$ 的最大整数)

简记 ${\displaystyle [a_n]=N}$, 则 ${\displaystyle N\le a_n<N+1}$. 根据递推式有$$\begin{array}{r}{a}_{n+1}-a_n=\frac{1}{a_n(a_n+1)}\in (\frac{1}{(N+1)(N+2)},\frac{1}{N(N+1)}],\end{array}$$
则考虑 ${\displaystyle [a_n]}$ 从第一次到达 ${\displaystyle N}$，到第一次到达 ${\displaystyle N+1}$ 所经历的指标的跨度，最少为 $$\begin{array}{r}\frac{1-{\displaystyle \frac{1}{N(N-1)}}}{{\displaystyle \frac{1}{N(N+1)}}}=N(N+1)-1+\frac{2}{N-1}>N(N+1)-1\end{array}$$
(也即走过的"路程"最短为 ${\displaystyle 1}$ 扣去上一个 ${\displaystyle N-1}$ 状态下的最大速度, 即 ${\displaystyle {\displaystyle 1-\frac{1}{(N-1)N}}}$, "速度" 最快为 ${\displaystyle {\displaystyle \frac{1}{N(N+1)}}}$), 且由于$$\begin{array}{r}\sum _{i=1}^6[i(i+1)-1]>100>\sum _{i=1}^5[i(i+1)-1],\end{array}$$
因此 ${\displaystyle a_{100}<6+a_1=7}$.

同理，这个跨度最大为 ${\displaystyle {\displaystyle \frac{1}{\frac{1}{(N+1)(N+2)}}=(N+1)(N+2)}}$, 且 $${\displaystyle \sum _{i=1}^5[(i+1)(i+2)]>100>\sum _{i=1}^4[(i+1)(i+2)],}$$ 因此 ${\displaystyle a_{100}>5+a_1=6}$.

综上，${\displaystyle [a_{100}]=6}$.