show whether $n2^n = 2^{O(n)}$ is true or not.In my opinion, it's false because O(n) can be n and thus the equality will be wrong, because $n2^n$ grows much faster than $2^n$
How to show whether $n2^n = 2^{O(n)}$?
asymptotics
One way, if you want to prove that $n2^n \in 2^{\mathcal{O}(n)}$, is to show that there exists a $c \in \mathbb{R}$ such that $n2^n \le 2^{cn}$ for all $n$ greater than some $n_0 \in \mathbb{N}$.Correspondingly, if you want to show the contrary, then you have to show that such a $c$ does not exist. However, we will see that the statement is true.Choose $c = 2$, i.e., we ask whether $n2^n \le^? 2^{2n} = 4^n$ for all $n > n_0$. We know that $n \le 2^n$ for all $n \in \mathbb{N}$. Therefore, it also holds that $n2^n \le 2^n\cdot 2^n = 2^{2n}$. Hence, we have an upper bound for the function $n2^n$ by choosing $c = 2$. In this case, we can set $n_0 = 1$, for instance.