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

To express the sequence \( \{a_n\}_{n=1}^{\infty} \) as an equivalent sequence starting from \( n=4 \), we first note the general term of the sequence is given by: \[ a_n = n^2 + 3n - 8 \] To find the equivalent sequence \( \{b_n\}_{n=4}^{\infty} \), we need to replace \( n \) with \( n + 3 \) to account for the new starting index. To do this: \[ b_n = a_{n+3} = (n+3)^2 + 3(n+3) - 8 \] Now, we can simplify this expression: 1. Expand \( (n+3)^2 \): \[ (n+3)^2 = n^2 + 6n + 9 \] 2. Expand \( 3(n+3) \): \[ 3(n+3) = 3n + 9 \] 3. Combine everything: \[ b_n = n^2 + 6n + 9 + 3n + 9 - 8 \] 4. Combine like terms: \[ b_n = n^2 + 9n + 10 \] Thus, the equivalent sequence is: \[ \{b_n\}_{n=4}^{\infty} = \{n^2 + 9n + 10\}_{n=4}^{\infty} \] This gives us the final result for the equivalent sequence.