Chapter 11 - Section 11.1 - Sequences and Summation Notation - Exercise Set - Page 829: 15

$4,12,48,240$

Need to find the first four terms of $a_n=2(n+1)!$ when $n=1,2,3,4$ Thus, $a_1=2(1+1)!=4$ $a_2=2(2+1)!=2.(3.2.1)=12$ $a_3=2(3+1)!=2(4.3.2.1)=48$ $a_4=2(4+1)!=2(5.4.3.2.1)=240$ Hence, the first four terms are: $4,12,48,240$