Determine the explicit formula for the \( n \)th term of the sequence below. \[ a_{1}=3, a_{n}=a_{n-1}+4^{n-2} \text { for } n \geq 2 \] Select the correct answer below: a \( a_{n}=3+n(4)^{n-2} \) \( a_{n}=3+n\left(4^{n-1}-1\right) \) \( a_{n}=3+\frac{4^{n}-1}{3} \) \( a_{n}=3+\frac{4^{n-1}-1}{3} \)

To derive the explicit formula for the sequence \( a_{1}=3 \) and \( a_{n}=a_{n-1}+4^{n-2} \), we can break down how each term relates to the previous one. For \( n=2 \), \( a_{2}=3+4^{0}=4 \), for \( n=3 \), \( a_{3}=4+4^{1}=8 \), and so forth. Observing this recursively, we recognize that the series is summing powers of 4 starting from \( 4^{0} \) through \( 4^{n-2} \). The last term in the sum will be \( 4^{n-2} \) with \( n \) terms total (starting from 1). This means we can express \( a_n \) as: \[ a_n = 3 + \sum_{k=0}^{n-2} 4^k = 3 + \frac{4^{n-1}-1}{3} \] Thus, the correct answer to choose is: \( a_{n}=3+\frac{4^{n-1}-1}{3} \). The joy of sequences lies in discovering sequences’ hidden patterns. This one specifically starts from a constant (3) and dynamically adds exponential growth increments, leading to towering terms as \( n \) increases! One might be surprised how easily one can overlook terms when deriving such patterns—always keeping meticulous track of the initial value and each recursive addition is essential in avoiding key mistakes along the way! Remembering the base case and systematically building up can transform complex recurrence into simple formulas!