Question or the following sequence, give an explicit formula for the \( n \)th term. Do not include " \( a_{n}= \) " in your response. \[ \frac{6}{7},-\frac{36}{13}, \frac{216}{19},-\frac{1296}{25}, \frac{7776}{31}, \ldots \]

Observe that the numerators of the terms are \[ 6,\ -36,\ 216,\ -1296,\ 7776,\ \dots \] These values can be expressed as powers of 6 with alternating signs. In particular, the first term is \( 6^1 \), the second is \(-6^2\), the third is \( 6^3 \), etc. A pattern of alternating signs emerges, which can be captured by the factor \((-1)^{n+1}\) when \( n \) starts at 1. Hence, the numerator is given by \[ (-1)^{n+1}6^n. \] Next, consider the denominators: \[ 7,\ 13,\ 19,\ 25,\ 31,\ \dots \] Notice that each successive term increases by 6. Thus, starting from 7, the \( n \)th denominator is \[ 7 + 6(n-1) = 6n + 1. \] Combining the two observations, an explicit formula for the \( n \)th term is \[ (-1)^{n+1}\frac{6^n}{6n+1}. \]