Exercise 2 Page 318

c|c a_1 =- 1, 5a_(n-1) & a_n = 6n-2 a_1=- 3, a_n = a_(n-1)+1 & a_1 = 9, a_n = 4a_(n-1) Recall that a rule can be explicit when it gives the nth term, a_n, as a function of the term's position number n. Or, it is recursive if it gives the first term or terms, and a recursive equation that tells how a_n is related to the preceding terms. Let's take a second look at the rules given and identify which of them are recursive rules. c|c a_1 =- 1, 5a_(n-1) & a_n = 6n-2 a_1=- 3, a_n = a_(n-1)+1 & a_1 = 9, a_n = 4a_(n-1) As we can see, all the rules are recursive since they give a first term and a recursive equation, except for a_n = 6n-2. This rule is explicit, as it just gives an equation for a_n in terms of the term's position. Therefore, this is the different one.