## Precalculus (6th Edition) Blitzer

To ﬁnd the first four terms of the sequence whose general term is $a_{n}$ =3$a_{n-1}$ - 1, we replace n in the formula with 2,3, and 4. n=1, $a_{1}$ = 5 n=2, $a_{2}$ =3$a_{2-1}$ -1= 3$a_{1}$ - 1 Substitute $a_{1}$ as 5 $a_{2}$ =3*5 -1 = 15 - 1= 14 n=3, $a_{3}$ =3$a_{3-1}$ - 1 = 3$a_{2}$ - 1 Substitute $a_{2}$ as 14 $a_{3}$ = 3*14 - 1= 42 - 1 = 41 n=4,$a_{4}$ =3$a_{4-1}$ - 1= 3$a_{3}$ - 1 Substitute $a_{3}$ as 41 $a_{4}$ =3*41 - 1 = 123 - 1 = 122 The ﬁrst four terms are 5, 14, 41,122.