We will use mathematical induction here. We can manually check that \(4n^2 + 4n\) is a multiple of 8 for \(n=0,1,2,3\). Let's replace \(n\) with \(n+1\):
\[4(n+1)^2 + 4(n+1)\]
Expand:
\[ \begin{gathered} 4(n^2 + 2n + 1) + 4n + 4 \\ 4n^2 +8n +4 +4n + 4 \\ (4n^2 +4n) + (8n +8) \end{gathered}\]
If (\(4n^2 + 4n\)) is divisible by 8, then \((4n^2 +4n) + (8n +8)\) is also divisible by 8, which means \(4(n+1)^2 + 4(n+1)\) is divisible by 8.