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

Question

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!

Answer 1

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}