Show that for any real constants \(a\) and \(b\), where \(b > 0\),

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

By the Binomial Theorem, \((n + a)^b = \sum\limits_{k=0}^{b}\binom{b}{k}n^{b-k}a{k} = C_{0}^{b}n^{b}a^0 + C_{1}^{b}n^{b-1}a^1 + \cdots + C_{b}^{b}n^{0}a^b\). In order to simplify further calculations, we want to restrict \(n + a\) to be non-negative which we can do by setting \(n_0 \geq \lvert a \rvert\). With \(n + a \geq 0\) we know that:

\[0 \leq C_{0}^{b}n^b \leq C_{0}^{b}n^{b}a^0 + C_{1}^{b}n^{b-1}a^1 + \cdots + C_{b}^{b}n^{0}a^b\]

We are interested in values of \(n\) such that \(n \geq n_0\), and since we know \(n_0\) is positive, we can use the fact that \(C_{0}^{b}n^{b}a^0 + C_{1}^{b}n^{b-1}a^1 + \cdots + C_{b}^{b}n^{0}a^b \leq (C_{0}^{b} + C_{1}^{b} + \cdots + C_{b}^{b})n^b\) to construct another step in our inequality:

\[0 \leq C_{0}^{b}n^b \leq C_{0}^{b}n^{b}a^0 + C_{1}^{b}n^{b-1}a^1 + \cdots + C_{b}^{b}n^{0}a^b \leq (C_{0}^{b} + C_{1}^{b} + \cdots + C_{b}^{b})n^b\]

Which is the definition of \(\Theta(n^b)\) with \(c_1 = C_{0}^{b} = 1\), \(c_2 = C_{0}^{b} + C_{1}^{b} + \cdots + C_{b}^{b}\) and \(n_0 \geq \lvert a \rvert\).