Answer
$\{n^2+6n-9\}_{n=1}^{\infty}=\{n^2+2n-17\}_{n=3}^{\infty}$
Work Step by Step
We are given the sequence:
$\{a_n\}=\{n^2+6n-9\}_{n=1}^{\infty}$
As we have to start the index of the sequence $\{b_n\}$ by 2 units upper than the sequence $\{a_n\}$, it means we must substitute $n-2$ for $n$ in $\{a_n\}$ to get $b_n$:
$b_n=(n-2)^2+6(n-2)-9$
$b_n=n^2-4n+4+6n-12-9$
$b_n=n^2+2n-17$
Therefore we have:
$\{n^2+6n-9\}_{n=1}^{\infty}=\{n^2+2n-17\}_{n=3}^{\infty}$
