Determine whether each of the functions $2^{n+1}$ and $2^{2n}$ is $O(2^n)$. Determine whether each of the functions $2^{n+1}$ and $2^{2n}$ is $O(2^n)$.
Since $2^n$ < $2^{n+1}$, you can say $2^{n+1}$ is  not $O(2^{n})$
Since $2^n$ is < $2^{2n}$, you can say  $2^{2n}$ is not $O(2^{n})$
Edit: more information:
Since $2^{n+1}$ < $3^{n+1}$ = (n+1) log 3 = O (n log3). 
Since $2^{2n}$ < $3^{2n}$ = O (2n log3) = O (n log3).

We know that $f(x) \in \mathcal{O}(g(x))$ if there exist constants $C$, and $k$ so that
$$ |f(x)| \le C |g(x)|$$
whenever $n > k$. I will omit the absolute value as it should be clear that $2^\text{r} \ge 0$.
(i) Now,
$$2^{n+1} = 2 \cdot 2^n $$
so we can take $C = 2, k=1$ to see that $2^{n+1} \in \mathcal{O}(2^n).$
(ii) Assume that $2^{2n} \in \mathcal{O}(2^n)$. Then, there exists $C$, $k$ so that
$$ 2^{2n} = 2^{n+n} =2^n2^n \le C \cdot 2^n \Rightarrow C \ge 2^n.$$
This is a contradiction as $C$ must be a constant and cannot depend on $n$. Hence, $2^{2n} \not\in \mathcal{O}(2^n).$
