## Precalculus (6th Edition) Blitzer

To ﬁnd the first four terms of the sequence whose general term is $a_{n}$ = $(-3)^{n}$. We replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = $(-3)^{1}$ = -3 n=2, $a_{2}$ =$(-3)^{2}$ = -3*-3 = 9 n=3, $a_{3}$ = $(-3)^{3}$ = -3*-3*-3 = -27 n=4,$a_{4}$ =$(-3)^{4}$ = -3*-3*-3*-3 = 81 The ﬁrst four terms are -3,9,-27 and 81.