Answer
$\{2n+1\}_{n=1}^{\infty}=\{2n-3\}_{n=3}^{\infty}$
Work Step by Step
We are given the sequence:
$\{a_n\}=\{2n+1\}_{n=1}^{\infty}$
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 $b_n$:
$b_n=2(n-2)+1$
$b_n=2n-4+1$
$b_n=2n-3$
Therefore we have:
$\{2n+1\}_{n=1}^{\infty}=\{2n-3\}_{n=3}^{\infty}$
