Is $$2^{n + 1} = O(2^n)$$? Is $$2^{2n} = O(2^n)$$?

#### Part 1 : Yes

Note that, $$2^{n + 1} = 2 \cdot 2^n$$

So, for $$n \ge 1$$ and any $$c \ge 2$$, $$0 \le 2^{n + 1} \le c \cdot 2^n$$.

Therefore, $$2^{n + 1} = O(2^n)$$.

#### Part 2 : No

Note that, $$2^{2n} = 2^n \cdot 2^n$$

Now, for $$2^{2n}$$ to be $$O(2^n)$$, we’ll need a constant $$c$$, such that $$0 \le 2^n \cdot 2^n \le c \cdot 2^n$$.

It is evident that we’ll need $$c \ge 2^n$$. But this is not possible for any arbitrarily large value of $$n$$. No matter what value of $$c$$ we choose, for some larger value $$n$$, it will not be sufficient.

So, $$2^{2n}$$ cannot be $$O(2^n)$$.