Exercise 57 Page 320

Practice makes perfect

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 .

a_{n} = a_{1} + (n - 1) d ⇓ a_{n - 1} = a_{1} + [(n - 1) - 1] d

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.

a_{n} = a_{1} + (n - 1) d

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

Associative Property of Addition

a_{n} = a_{1} + [((n - 1) - 1) + 1] d

Distribute $d$

a_{n} = a_{1} + [(n - 1) - 1] d + d

$a_{1} + [(n - 1) - 1] d = a_{n - 1}$

a_{n} = a_{n - 1} + d

We have showed that the recursive rule for a sequence can be obtained by manipulating the explicit rule.

Communicate Your Answer p.313

Monitoring Progress p.315-317

Exercises p.318-320

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