For each sequence, find a formula for the general term, \( a_{n} \). For example, answer \( n^{2} \) if given the sequence: \( \{1,4,9,16,25,36, \ldots\} \) \begin{tabular}{|}\( \square \) & \( \left\{\frac{3}{16}, \frac{4}{25}, \frac{5}{36}, \frac{6}{49}, \ldots\right\} \) \\ \( \left.\square \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots\right\} \)\end{tabular}

**Sequence 1** Consider the sequence: \[ \left\{\frac{3}{16}, \frac{4}{25}, \frac{5}{36}, \frac{6}{49}, \ldots \right\} \] - **Step 1:** Identify the numerator. The numerators are: 3, 4, 5, 6, ... This pattern suggests that the numerator for the \( n \)-th term is \( n+2 \). - **Step 2:** Identify the denominator. The denominators are: 16, 25, 36, 49, ... Notice that: \[ 16=4^2,\quad 25=5^2,\quad 36=6^2,\quad 49=7^2, \ldots \] Thus, the denominator for the \( n \)-th term is \((n+3)^2\). - **Step 3:** Write the formula for the \( n \)-th term. Combining the observations: \[ a_n = \frac{n+2}{(n+3)^2} \] --- **Sequence 2** Consider the sequence: \[ \left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots\right\} \] - **Step 1:** Identify the numerator. The numerators are: 1, 2, 3, 4, ... Clearly, the numerator for the \( n \)-th term is \( n \). - **Step 2:** Identify the denominator. The denominators are: 2, 3, 4, 5, ... Clearly, the denominator for the \( n \)-th term is \( n+1 \). - **Step 3:** Write the formula for the \( n \)-th term. Combining the observations: \[ a_n = \frac{n}{n+1} \]