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Algebra 2 Honors · Lesson 5.2

Explicit and Recursive Rules

How it works

An explicit rule gives a term directly from its position. A recursive rule gives an initial term and a method for obtaining later terms from previous ones. A recursive rule without a starting value generally does not specify a unique sequence.

an=an−1+d(n≥2),a1 specifieda_n=a_{n-1}+d\quad(n\ge2),\qquad a_1\text{ specified}

Worked examples

Example 1

Write explicit and recursive rules for 5,9,13,…\displaystyle 5,9,13,\ldots.

The difference is 4\displaystyle 4. Explicit: an=5+4(n−1)\displaystyle a_n=5+4(n-1). Recursive: a1=5\displaystyle a_1=5, an=an−1+4\displaystyle a_n=a_{n-1}+4 for n≥2\displaystyle n\ge2.

Example 2

Given a1=12\displaystyle a_1=12 and an=an−1/2\displaystyle a_n=a_{n-1}/2, find a4\displaystyle a_4.

The terms are 12,6,3,3/2\displaystyle 12,6,3,3/2, so a4=3/2\displaystyle a_4=3/2.

Problems

Problem 5.2.1

Write a recursive rule for 2,6,18,…\displaystyle 2,6,18,\ldots.

Show worked solution

a1=2\displaystyle a_1=2 and an=3an−1\displaystyle a_n=3a_{n-1} for n≥2\displaystyle n\ge2.

Problem 5.2.2

Write an explicit rule for a1=7\displaystyle a_1=7, an=an−1−2\displaystyle a_n=a_{n-1}-2.

Show worked solution

The arithmetic difference is −2\displaystyle -2, so an=7−2(n−1)\displaystyle a_n=7-2(n-1).

Problem 5.2.3

Given b1=3\displaystyle b_1=3, bn=2bn−1+1\displaystyle b_n=2b_{n-1}+1, find b2,b3,b4\displaystyle b_2,b_3,b_4.

Show worked solution

Apply the rule sequentially: b2=7\displaystyle b_2=7, b3=15\displaystyle b_3=15, and b4=31\displaystyle b_4=31.

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