Q:

Proof by induction of $s_k=2s_{k-2}$

so the question is:
Question: Guess the explicit formula and provide a mathematical induction as proof.
$s_k=2s_{k-2}$
for all integers $k\geq2$
$ s_0=1, s_1=2 $
I have  the following figured out:
$s_2=2$, $s_3=4$, $s_4=4$, $s_5=8$, $s_6=8$
$2^0,2^1,2^1,2^2,2^2,2^3,2^3...$
For even numbers the explicit formula is: $2^\frac{n}{2}$
And for odd numbers, the explicit formula is: $2^\frac{n+1}{2}$
Thus, $2^{\left \lfloor{\frac{n+1}{2}}\right \rfloor }$ is the explicit formula.
Following someone's previous advice, I have been told that proof by induction where P(k) is assumed to be P(k+2) is more straightforward because we are given $s_k=2s_{k-2}$.
Induction:
Base Step: 
$P(1)=2^\frac{1+1}{2}=2^1=2$
$P(2)=2^\frac{2}{2}=2^1=2$
Inductive step:
Case 1: if $P(k)=2^\frac{k+1}{2}=2s_{k-2}$
$P(k+2)=2^\frac{k+1+2}{2}=2^\frac{k+3}{2}$
$S_{k+2}=2s_{k-2+2}=2s_k$
$P(k+2)= 2^\frac{k+3}{2}=2s_k$
$2^\frac{k+3}{2} = 2^\frac{k+1}{2} * 2^\frac{2}{2}$
$=2s_{k-2} + 2=2s_k$
And from there on I have no idea what to do. Please help! Thank you!

A:

Assuming$$P(k): s_k = 2^{\lfloor \frac{k+1}2 \rfloor} $$
we want to prove that
$$P(k+2): s_{k+2}=2^{\lfloor \frac{k+3}2 \rfloor}$$
We start from LHS
\begin{align}
LHS &=s_{k+2} \\&= 2s_k \\
&= 2\cdot2^{\lfloor  \frac{k+1}{2}\rfloor} \\
&= 2^{\lfloor  \frac{k+1}{2}\rfloor+1} \\
&= 2^{\lfloor  \frac{k+3}{2}\rfloor}\\
&= RHS
\end{align}

