Exercise 3.1.2

Show that for any real constants $a$ and $b$, where $b > 0$,

$$(n + a)^b = \Theta(n^b)$$

Quite simply:

$$(n + a)^b = \binom{n}0n^b + \binom{n}1n^{b-1}b + \cdots + \binom{n}0a^b$$

The most significant term is $n^b$ and this is obviously polynomially tightly bound.