Show that for any real constants $a$ and $b$, where $b > 0$, $(n + a)^b =\Theta(n^b)$.

To prove this, we have to show that there exists constants $c_1, c_2, n_0 > 0$ such that $0 \le c_1 n^b \le (n + a)^b \le c_2 n^b$ for all $n \ge n_0$.

Note that, $n + a \le 2n$, when $|a| \le n$ Also note, $n + a \ge \frac 1 2 n$, when $|a| \le \frac n 2$

Therefore, when $n \ge 2 \vert a \vert$,

As $b > 0$, we can raise all the terms of the previous inequality to the power of $b$ without breaking the inequality:

So, $(n + a)^b = \Theta(n^b)$ because there exists $c_1 = 1/{2^b}$, $c_2 = 2^b$, and $n_0 = 2 \vert a \vert$.