Evaluate the product \(\prod_{k = 1}^n 2 \cdot 4^k\).

Let’s assume \(P = \prod_{k = 1}^n 2 \cdot 4^k\)

\[\begin {align} \lg P & = \lg \left( \prod_{k = 1}^n 2 \cdot 4^k \right) \\ & = \sum_{k = 1}^n \lg \left(2 \cdot 4^k \right) \\ & = \sum_{k = 1}^n \lg \left(2 \cdot 2^{2k} \right) \\ & = \sum_{k = 1}^n \lg 2^{2k + 1} \\ & = \sum_{k = 1}^n (2k + 1) \\ & = 2\sum_{k = 1}^n k + \sum_{k = 1}^n 1 \\ & = 2 \frac {n(n + 1)} 2 + n \\ & = n(n + 1) + n \\ & = n(n + 2) \end {align}\]Hence, \(P = 2^{n(n + 2)}\).