The question asks:
Let $a_n$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n$ + 5 .
Use induction to prove that $a_n \gt 2^n$ for all $n \in \mathbb N$ .
I proved the base case and for the induction step I proved it like so:
\begin{align} a_{k+1} &\gt 2^{k+1}\\ 2a_{k}+5 &> 2^{k+1} \tag{from the original problem}\\ a_{k} &> 2^{k} - (5/2) \end{align}
and because
$$2^k \gt 2^{k} - \frac{5}{2}$$
I finalized my proof.
Is this correct?
Your induction step, as I commented, is incorrect because you have started with the very statement you are trying to prove! To avoid this mistake, either start with the LHS or RHS of the equation to deduce the final statement.
Don't forget the inductive assumption is $a_{k}\gt 2^k$.
Your proof might look something like this:
\begin{align} LHS & = a_{k+1}\\ &=2a_k +5 \\ & \gt2\cdot2^k+5 \tag{by the inductive assumption}\\ & =2^{k+1}+5\\ &\gt2^{k+1}\\ &= RHS \end{align}
$\therefore LHS \gt RHS \implies a_{k+1}\gt 2^{k+1}$. The inductive step is done so the induction is completed.