I need to prove that $2^{n+1} = \Theta(2^n)$?
To do this, I need to show that there are positive constants $c_1,c_2$ such that
$$c_1 2^n \le 2^{n + 1} \le c_2 2^n$$ ...for all sufficiently large $n$.
To find $c_2$, I know that I just need to find $O(2^n)$ which I have done below:
$$ \frac{f(n)}{g(n)} = \frac{2^{n+1}}{2^n}=\frac{2 \cdot {2^n}}{2^n} = 2 $$
Therefore, $c_2 = 2$.
Can anybody tell me how to find $c_1$?
There are several choices available, but here's a hint: What number, when multiplied to $2^n$, will give you $2^{n+1}$?