Let ${a_n}$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n + 5$. Use induction to prove that $a_n > 2^n$ for all $n \in N$ Let ${a_n}$ be the sequence given by $a_1 = 3$ and $a_{n+1} = 2a_n + 5$. Use induction to prove that $a_n > 2^n$ for all $n \in \mathbb{N}$
Attempt:
For $n = 1,$ we have  $3>2$ so the base case holds.
Assume  $a_k > 2^k$ for some k
For $n = k+1$ we have:
$a_{k+1} = 2 a_{k} + 5$

You could infact obtain $a_n$ with reasonable ease; Note that
$$a_{n+1} + 5 = 2(a_n+5)$$
Hence, we have
$$a_{n+1} + 5 = 2(a_n+5) = 2(2(a_{n-1}+5)) = 2^k (a_{n-k+1}+5) = 2^n(a_1 + 5)$$
Hence, we get
$$a_{n+1} +5 = 2^n(3+5) = 2^{n+3} \implies a_n = 2^{n+2} - 5$$
Now show that for $n \geq 1$, we have
$$2^{n+2} - 5 > 2^n$$
