Recall that $Z^{+}=\{1,2,3,$dots$\}.$ Consider the following recursive algorithm mystery:

Algorithmmystery$(n)$ : $,($assume $($:$nin{Z}^{+}\}$

If $n=1$ or $n=2$ then return $n$

else return mystery $(n-1)+2**$ mystery $(n-2)$

Here are examples of how mystery would exectue on a few first inputs $nin{Z}^{+}$:

mystery$(1)$ returns $1$ by the base case,

mystery$(2)$ returns $2$ by the base case,

mystery$(3)$ returns mystery$(2)+2**$ mystery $(1)=2+2**1=4,$

mystery$(4)$ returns mystery$(3)+2**$ mystery $(2)=[dots]$

Do two things: $(1)$ state a closed$-$form formula for function $f(n)$ s$.$t$.f(n)$ is the output of mystery $(n),$ for every $nin{Z}^{+},$ and $(2)$ use induction to prove that mystery $(n)=f(n)$ for all $n.{?}^1$