Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1048: 21

4,12,48,240

To find the first four terms of the sequence whose general term is $a_{n}$ = 2(n+1)!. We replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = 2(1+1)!= 2*2 = 4 n=2, $a_{2}$ =2(2+1)! = 2*3! =2*3*2 = 12 n=3, $a_{3}$ = 2(3+1)! = 2*4! =2*4*3*2 = 48 n=4,$a_{4}$ =2(4+1)! = 2*5! =2*5*4*3*2 = 240 The first four terms are 4,12,48,240