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

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$t_{1} = \frac{6 ( 1 + 1 )!}{3 ( 1 )} = \frac{2 ( 2 )!}{1} = 4$

$t_n=\dfrac{6(n+1)!}{3n}=\dfrac{2(n+1)!}{n}, n\geq1$

$t_{n+1}=\dfrac{2(n+1+1)!}{n+1}=\dfrac{2(n+1)!(n+2)}{n+1}$

$=\dfrac{2(n+1)!}{n}(\dfrac{n(n+2)}{n+1})=t_n(\dfrac{n(n+2)}{n+1})$

$t_{n+1}=t_n\cdot\dfrac{n(n+2)}{n+1}, n\geq1$