Exercise 57 Page 320

Practice makes perfect

We know that the explicit rule defines an arithmetic sequence. a_n=a_1+(n-1)d When we substitute ( n-1) for n in the explicit rule, we get a_(n-1)=a_1+[(n-1)-1]d. \begin{gathered} a_n=a_1+(n-1)d \\ \Downarrow \\ a_{{\color{#0000FF}{n-1}}}=a_1+[({\color{#0000FF}{n-1}})-1]d \end{gathered

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<listcircle icon="b"> We will use the equation in Part A when showing the recursive equation. Before that, we need to generate a_1+[(n-1)-1]d on the right hand side of the explicit rule. </listcircle>

a_n=a_1+(n-1)d

IdPropAdd

Identity Property of Addition

a_n=a_1+[(n-1)+0]d

Rewrite 0 as (- 1)-(- 1)

a_n=a_1+[(n-1)-1+1)]d

AssociativePropAdd

Associative Property of Addition

a_n=a_1+[((n-1)-1)+1]d

Distr

Distribute d

a_n=a_1+[(n-1)-1]d+d