Let us find a recursive definition of the sequence with general term
tn=3n6(n+1)!,n≥1.
It follows that
tn+1=3n+16(n+2)!=3n⋅36(n+1)!(n+2)=3n6(n+1)!3n+2=tn3n+2.
We conclude that the recursive definition of this sequence is
tn+1=31(n+2)tn, t1=4.
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