Find, showing all working, a recursive definition of the sequence with general term tn = 6 (n + 1)!/3n, n >= 1

Let us find a recursive definition of the sequence with general term

$t_n = \frac{6 (n + 1)!}{3^n}, n \ge 1.$

It follows that

$t_{n+1} = \frac{6 (n + 2)!}{3^{n+1}}=\frac{6 (n + 1)!(n+2)}{3^{n}\cdot3}=\frac{6 (n + 1)!}{3^{n}}\frac{n+2}{3}=t_n\frac{n+2}{3}.$

We conclude that the recursive definition of this sequence is

$t_{n+1}=\frac{1}3(n+2)t_n,\ t_1=4.$