Express the sequence \( \left\{a_{n}\right\}_{n=1}^{\infty} \) as an equivalent sequence of the form \( \left\{b_{n}\right\}_{n=4}^{\infty} \). \( \left.\qquad n^{2}+3 n-8\right\}_{n=1}^{\infty} \) An equivalent sequence is \( \left\}_{n=4}^{\infty}\right. \). (Simplify your answer. Do not factor.)

To express the sequence \( \{a_n\}_{n=1}^{\infty} = n^2 + 3n - 8 \) as an equivalent sequence \( \{b_n\}_{n=4}^{\infty} \), we need to find a new sequence starting from \( n=4 \). First, we'll replace \( n \) with \( m + 3 \) where \( m \) starts from 1 (so \( n = m + 3 \) with \( m = 1 \) yielding \( n = 4 \), and as \( m \) increases, \( n \) will cover values starting from 4). Thus: \[ a_n = (m + 3)^2 + 3(m + 3) - 8 \] Now, let's simplify this expression: \[ (m + 3)^2 = m^2 + 6m + 9 \] \[ 3(m + 3) = 3m + 9 \] Combining these gives: \[ a_n = (m^2 + 6m + 9) + (3m + 9) - 8 \] \[ = m^2 + 6m + 9 + 3m + 9 - 8 \] \[ = m^2 + 9m + 10 \] So, we have: \[ b_m = m^2 + 9m + 10 \] Now, since \( n = m + 3 \), we can say they’re equivalent: \[ \{b_n\}_{n=4}^{\infty} = \{m^2 + 9m + 10\}_{m=1}^{\infty} \] Thus, the equivalent sequences are: \[ b_n = (n - 3)^2 + 9(n - 3) + 10 \] for \( n \geq 4 \).