\(a_{n-1} = (n-1)^2 + 3(n-1) + 4 = n^2 + n + 2\text{.}\) Thus \(a_n - a_{n-1} = 2n+2\text{.}\) Note that this is linear (arithmetic). We can check that we are correct. The sequence \(a_n\) is \(4, 8, 14, 22, 32, \ldots\) and the sequence of differences is thus \(4, 6, 8, 10,\ldots\) which agrees with \(2n+2\) (if we start at \(n = 1\)).