Let $A=\{{5}^n-4n-1\colon n\in N\}$ and $B=\{{}16(n-1)\colon n\in N\}$ be sets. Then

We are given two sets: \[ A = \{\,5^n - 4n - 1 : n \in \mathbb{N}\,\} \] \[ B = \{\,16(n-1) : n \in \mathbb{N}\,\} \] We test the first few values. For set \(A\): \[ \begin{aligned} n=1 &: 5^1 - 4 - 1 = 0 \\ n=2 &: 25 - 8 - 1 = 16 \\ n=3 &: 125 - 12 - 1 = 112 \\ n=4 &: 625 - 16 - 1 = 608 \end{aligned} \] So, \[ A = \{0,\;16,\;112,\;608,\dots\} \] For set \(B\): \[ \begin{aligned} n=1 &: 16(0) = 0 \\ n=2 &: 16(1) = 16 \\ n=3 &: 16(2) = 32 \\ n=4 &: 16(3) = 48 \end{aligned} \] So, \[ B = \{0,\;16,\;32,\;48,\;64,\dots\} \] Clearly every element of \(A\) also appears in \(B\), since: \[ 5^n - 4n - 1 = 16(n-1) \] Hence: \[ A \subset B. \] Therefore, the correct answer is: \[ \boxed{A \subset B} \]