The sequence $\{a_n\}, n \in N$ satisfies $a_1 = 1$ and $5^{a_{n+1}-a_n} = 1 + \frac{1}{n + \frac{2}{3}}$. Then: (where $[.]$ denotes greatest integer function)
Step-by-Step Solution
Key Concept: Use telescoping products to find the general term, then solve logarithmic inequalities to find the range of $n$.
From $\frac{5^{6n+1}}{5^{6n}} = \frac{3n+5}{3n+2}$, telescoping the product $\frac{5^{n_2}}{5^{n_1}} \cdot \frac{5^{n_3}}{5^{n_2}} \cdots \frac{5^{n_n}}{5^{n_{n-1}}}$ gives $\frac{5^{n_n}}{5^{n_1}} = \frac{3n+2}{5}$, so $5^{n_n} = 3n+2$, thus $a_n = \log_5(3n+2)$. For $|a_n| = 3$, we get $\log_5(3n+2) \in [3,4]$, giving $n \in [41, 42, \ldots, 207]$ and $n \in [208, 209, \ldots, 1041]$ for $[a_n] = 3$ and $[a_n] = 4$ respectively.
Correct Answer: 1,2,4