Prove by induction $s_k = 2s_{k−2}$, for all integers $k \le 2, s_0 = 1,s_1 = 2$ has the explicit formula $s_n = 2^{\lceil\frac{n}{2}\rceil}.$
I am only intersted in the inductive step.
Let $s_i = 2^{\lceil\frac{i}{2}\rceil}$ be the hypothesis which is valid for all integers $i$ with $0 \le i \le k.$
Then $s_{k + 1} = 2s_{k−1} = 2\cdot 2^{\lceil\frac{i - 1}{2}\rceil} = 2^{1 +\lceil\frac{i - 1}{2}\rceil }$. I am kind of stuck here.
Since ${\lceil1\rceil} = 1$ we should have ${\lceil1\rceil} + \lceil\frac{i - 1}{2}\rceil = \lceil\frac{i - 1}{2} +1\rceil = \lceil\frac{i + 1}{2}\rceil$, but I am not sure if this kind of manipulation is allowed. Does it make sense to say $s_{k + 1} = 2^{1 +\lceil\frac{i - 1}{2}\rceil } = 2^{\lceil\frac{i + 1}{2}\rceil}$?
Since you’re really dealing with two unrelated sequences, one for the terms with even index and one for the terms with odd index, it would be much easier to use two separate inductions and then combine the results at the end.
First prove by induction on $n$ that $s_{2n}=2^n$ for $n\ge 0$. This is straightforward. For the basis step we have $s_{2\cdot0}=s_0=1=2^0$, and if $s_{2n}=2^n$, then $s_{2(n+1)}=2s_{2n}=2\cdot2^n=2^{n+1}$.
Then prove by induction on $n$ that $s_{2n+1}=2^{n+1}$. This is equally straightforward, and I’ll leave the details to you.
Then simply observe that
$$\left\lceil\frac{2n}2\right\rceil=n$$
and
$$\left\lceil\frac{2n+1}2\right\rceil=n+1\;,$$
so that in both cases $s_n=2^{\lceil n/2\rceil}$.