\(^{n+1}C_3 = 4(^nC_3)\\\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\\\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)\\=\frac{n+1}{n-2}=\frac{4}{1}\\n+1 = 4(n-2)\\n+1 = 4n-8\\-3n = -9\\\frac{-9}{-3}\\n=3\)