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.)

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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 way to re-index the sequence starting from $$ n=4 $$.

1. **Identify the original sequence**: 
   $$
   a_n = n^2 + 3n - 8
   $$

2. **Determine the new index**: 
   We want to express $$ b_n $$ in terms of $$ a_n $$ but starting from $$ n=4 $$. To do this, we can set $$ m = n - 3 $$ so that when $$ n = 4 $$, $$ m = 1 $$.

3. **Substitute the new index**: 
   We can express $$ n $$ in terms of $$ m $$:
   $$
   n = m + 3
   $$
   Now, substitute $$ n $$ into the expression for $$ a_n $$:
   $$
   b_m = a_{m+3} = (m+3)^2 + 3(m+3) - 8
   $$

4. **Simplify the expression**:
   $$
   b_m = (m^2 + 6m + 9) + (3m + 9) - 8
   $$
   Combine like terms:
   $$
   b_m = m^2 + 6m + 9 + 3m + 9 - 8
   $$
   $$
   b_m = m^2 + 9m + 10
   $$

5. **Final expression**:
   Since $$ m $$ starts from 1 when $$ n = 4 $$, we can express the equivalent sequence as:
   $$
   b_n = n^2 + 9n + 10 \quad \text{for } n \geq 1
   $$

Thus, the equivalent sequence is:
$$
\{b_n\}_{n=4}^{\infty} = n^2 + 9n + 10
$$
The equivalent sequence is $$ b_n = n^2 + 9n + 10 $$ for $$ n \geq 4 $$.

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.)

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