In first case, $c \geq 2$, So, we can find atleast one positive real 'c' for which definition of Big O will be satisfied.. Right?

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

- **Yes. **

**- Reason : **$\space 2 \cdot 2^{n} \le c \cdot 2^{n}$ where $c$ is a positive real number, $c \ge 2$ in this particular case.

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

- **No. **

**- Reason : **

$2^{n+n} \le c \cdot 2^{n} $ where c is a positive real number. There's no such constant $c$ that satisfies this inequality.

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