Given that \(^{n+1}C_{3} = 4 \cdot ^{n}C_{3}\), we can write this as:

\(\frac{(n+1)!}{3!(n+1-3)!} = 4 \cdot \frac{n!}{3!(n-3)!}\)

Simplifying both sides of the equation:

\(\frac{(n+1)(n)(n-1)(n-2)}{6} = 4 \cdot \frac{n(n-1)(n-2)}{6}\)

Now, we can cancel out the common factors:

\((n+1) = 4\)

So, \(n = 3\).

Therefore, the answer is 3