Explicit against recursive rules, generating terms from a recursive rule, writing recursive rules for arithmetic and geometric sequences and for the Fibonacci and factorial sequences, recursive models with long-run behaviour, and iterating a function.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 12 — Sequences and Series
Use Recursive Rules with Sequences and Functions
Objectives
Five outcomes. A different way to say what a sequence is.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 827-831 — the lesson these objectives are drawn from
Warm-up
Every rule so far in this chapter has been explicit: a_n written directly in terms of n.
Discussion prompt
The sequence 1, 5, 9, 13, 17 has the explicit rule a_n = 1 + 4n starting at n equal to 0. Describe the same sequence WITHOUT mentioning n at all.
Hint: Say where it starts and what each step does.
Answer:
Start at 1, and each term is 4 more than the one before.
\[ a_0 = 1, \quad a_n = a_{n-1}+4 \]
That is a recursive rule: a starting term plus an equation saying how each term relates to the ones before it. It describes the same list from a completely different angle, and for some sequences it is the only description that is easy to write.
Concept
An explicit rule gives a_n as a function of the position n. A recursive rule gives the starting term or terms and an equation building each term from the ones before it. Both can describe the same sequence, and each is easier for different sequences.
recursive rule — A rule that gives the beginning term or terms of a sequence together with a recursive equation telling how a_n is related to one or more preceding terms. Both halves are required.
\[ \text{arithmetic: } a_n = a_{n-1}+d; \qquad \text{geometric: } a_n = r\,a_{n-1} \]
The Fibonacci and factorial sequences have simple recursive rules and awkward explicit ones, which is the clearest argument for the idea.
Figure (svg): Two columns comparing explicit and recursive rules
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 827-828
Section
Section 1
Concept
To generate terms from a recursive rule, write down the given starting term, then apply the recursive equation once for each further term, always using the value you just produced.
\[ a_0 = 1, \; a_n = a_{n-1}+4 \;\Longrightarrow\; 1,5,9,13,\dots \]
There is no shortcut. Reaching the fiftieth term means computing all forty-nine before it, which is the price of the recursive form.
Figure (svg): A recursive rule generating terms one link at a time
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 827-827 — Evaluate recursive rules
Picture it
Example 1a: adding four each step.
Figure (svg): A recursive rule generating terms one link at a time
Each box is built from the one to its left. The explicit rule at the bottom describes the identical list but reaches any term at once.
Worked example
Example 1, both parts.
\[ \text{List six terms of } a_0 = 1, a_n = a_{n-1}+4 \text{ and of } a_1 = 1, a_n = 3a_{n-1}. \]
First: start at the given term
Why: A sub zero is 1.
\[ 1 \]
First: apply the equation five times
Why: Add 4 each step.
\[ 1, 5, 9, 13, 17, 21 \]
Second: start at the given term
Why: A sub one is 1.
\[ 1 \]
Second: apply the equation five times
Why: Multiply by 3 each step.
\[ 1, 3, 9, 27, 81, 243 \]
Figure (svg): A recursive rule generating terms one link at a time
\[ 1,5,9,13,17,21; \quad 1,3,9,27,81,243 \]
Verify: classify each sequence
Why: The first has constant differences of 4, so it is arithmetic; the second has constant ratios of 3, so it is geometric. That is not an accident — adding a constant in the recursive equation always gives arithmetic, and multiplying by one always gives geometric.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 827-827
Fill the middle
Example 1b.
Fill in the blanks
a_1 = 1, \; a_n = 3a_27 \;\Longrightarrow\; a_4 = 3a_3 = 3(9) = ___
Why: Three times the third term, 9, gives 27. Each application uses the value produced by the last one.
Worked example
Guided Practice 1 to 4.
\[ \text{List five terms of each: } a_1 = 3, a_n = a_{n-1}-7; \; a_0 = 162, a_n = 0.5a_{n-1}; \; a_0 = 1, a_n = a_{n-1}+n; \; a_1 = 4, a_n = 2a_{n-1}-1. \]
First: subtract 7 each step
Why: Three, then negative 4, and so on.
\[ 3, -4, -11, -18, -25 \]
Second: halve each step
Why: One sixty-two, then 81.
\[ 162, 81, 40.5, 20.25, 10.125 \]
Third: add the position number
Why: One, then 1 plus 1, then 2 plus 2.
\[ 1, 2, 4, 7, 11 \]
Fourth: double and subtract 1
Why: Four, then 7, then 13.
\[ 4, 7, 13, 25, 49 \]
Figure (svg): The solution to Worked example four more rules shown as a ladder of expressions, one row per algebraic move
\[ 3,-4,-11,\dots; \; 162,81,40.5,\dots; \; 1,2,4,7,11; \; 4,7,13,25,49 \]
Verify: classify the third and fourth
Why: The third has differences 1, 2, 3, 4 — growing, so it is neither arithmetic nor geometric. The fourth has differences 3, 6, 12, 24, which double, so it is neither either. A recursive equation that mixes operations, or that mentions n, escapes both families.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 828-828
Trap
\[ a_0 = 1, \; a_n = a_{n-1}+4 \]
Substitute n equal to 3 directly
Why: The equation is treated as if it were explicit.
\[ a_3 = 3-1+4 = 6 \quad \text{(wrong)} \]
The symbol a sub n minus 1 is the PREVIOUS TERM, not the number n minus 1. It is a term of the sequence, not an arithmetic expression.
\[ a_1 = 1+4 = 5, \; a_2 = 5+4 = 9, \; a_3 = 9+4 = 13 \]
Work forward one term at a time
Why: Each application uses the value just computed.
\[ a_3 = 13 \]
Reading a sub n minus 1 aloud as the term before this one, rather than as n minus 1, prevents the whole class of error.
Matching
Start at the given term and step.
Match the pairs
Why: The third and fourth start out looking like doubling but are not: 1, 2, 4 then 7 breaks the pattern, and 4, 7, 13 was never doubling to begin with. Generating four or five terms is the only reliable way to see what a recursive rule does.
Sorting
Look at the shape of the recursive equation.
Sort into buckets
Sort each recursive equation by the kind of sequence it generates.
The last shape, multiply then add, is the one Example 4 uses to model a real population, and it is exactly the shape that settles toward a stable level.
Prediction
Commit before reasoning.
Predict first
The sequence 1, 5, 9, 13, ... has both an explicit and a recursive rule. Which finds a_50 with less work?
Correct: The explicit rule, in one substitution; the recursive one needs fifty steps.
\[ a_{50} = 1+4(50) = 201 \]
Why: The explicit rule 1 plus 4n gives 201 immediately at n equal to 50, while the recursive rule must produce every term in between. That is the price of the recursive form, and it is worth paying only when the recursive rule is much easier to WRITE — as it is for the Fibonacci sequence, where no simple explicit rule presents itself at all.
Section
Section 2
Concept
For an arithmetic sequence the recursive equation adds the common difference; for a geometric one it multiplies by the common ratio. Either way the rule must also give the first term.
\[ a_n = a_{n-1}+d; \qquad a_n = r\,a_{n-1} \]
A recursive equation without an initial term describes infinitely many different sequences, so the starting value is part of the rule, not an extra.
Figure (svg): The recursive equations for arithmetic and geometric sequences
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 827-828 — Recursive Equations for Arithmetic and Geometric Sequences
Picture it
The two standard shapes, and what a complete rule must contain.
Figure (svg): The recursive equations for arithmetic and geometric sequences
The equation says how to step; the initial term says where to start. Leave out either and the sequence is not determined.
Worked example
Example 2, both parts.
\[ \text{Write recursive rules for } 3,13,23,33,43,\dots \text{ and } 16,40,100,250,625,\dots \]
First: test for a common difference
Why: Thirteen minus 3 is 10, and so on.
First: write both halves
Why: The first term and the step.
\[ a _{1} = 3, a _{n} = a _{n - 1} + 10 \]
Second: test for a common ratio
Why: Forty over 16 is 2.5, and so on.
Second: write both halves
Why: The first term and the step.
\[ a _{1} = 16, a _{n} = 2.5 a _{n - 1} \]
Figure (svg): The recursive equations for arithmetic and geometric sequences
\[ a_1 = 3, \, a_n = a_{n-1}+10; \qquad a_1 = 16, \, a_n = 2.5a_{n-1} \]
Verify: run each rule forward two steps
Why: The first gives 3, 13, 23 and the second gives 16, 40, 100, matching the sequences as printed. Generating two or three terms from a rule you just wrote is a five-second check that catches a wrong d or r immediately.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 828-828
Fill the middle
Example 2b.
Fill in the blanks
16, 40, 100, 250 \;\Longrightarrow\; r = \tfrac2.5___ = ___
Why: Forty over 16 is 2.5, and checking 100 over 40 gives 2.5 as well. The recursive equation is therefore a_n equals 2.5 times the previous term.
Worked example
Guided Practice 5 to 8.
\[ \text{Write recursive rules for } 2,14,98,686; \; 19,13,7,1; \; 11,22,33,44; \; 324,108,36,12. \]
First: fourteen over 2
Why: A constant ratio of 7.
\[ a _{1} = 2, a _{n} = 7 a _{n - 1} \]
Second: thirteen minus 19
Why: A constant difference of negative 6.
\[ a _{1} = 19, a _{n} = a _{n - 1} - 6 \]
Third: twenty-two minus 11
Why: A constant difference of 11.
\[ a _{1} = 11, a _{n} = a _{n - 1} + 11 \]
Fourth: one hundred eight over 324
Why: A constant ratio of one third.
\[ a _{1} = 324, a _{n} = (\frac{1}{3}) a _{n - 1} \]
Figure (svg): The solution to Worked example four more rules shown as a ladder of expressions, one row per algebraic move
\[ 7a_{n-1}; \; a_{n-1}-6; \; a_{n-1}+11; \; \tfrac{1}{3}a_{n-1} \]
Verify: watch the third one carefully
Why: The terms 11, 22, 33, 44 double from the first to the second, which tempts a ratio of 2 — but 33 over 22 is 1.5, not 2. The differences are all 11, so it is arithmetic. This is the sequence in this set most likely to be misclassified from its first two terms alone.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 828-828
Error analysis
A student writes a recursive rule for 3, 13, 23, 33, 43.
Annotate
On: \( a_n = a_{n-1}+10 \)
The textbook prints this caution on page 828. A recursive EQUATION is half a rule; the initial term is the other half.
Matching
Test differences first, then ratios.
Match the pairs
Why: Two are arithmetic and two geometric, and each rule carries its own initial term. Notice how much shorter these are than the explicit rules for the same four sequences.
Comparison
Fill the blanks. The same sequence 3, 13, 23, 33.
Comparison matrix
| Question | Explicit | Recursive |
|---|---|---|
| Rule | a_n = 10n - 7 | a_1 = 3, a_n = a_(n-1) + 10 |
| Finding a_100 | one substitution | ninety-nine steps |
| What it describes | the term at a position | the step between terms |
| Needs a starting term? | no | yes |
For a plain arithmetic sequence the explicit rule wins on convenience. The recursive form earns its keep when the process is easy to describe but the position formula is not.
Prediction
Commit before reasoning.
Predict first
How many sequences satisfy the equation a_n = a_(n-1) + 10 with no initial term given?
Correct: Infinitely many — one for every possible starting value.
\[ 3,13,23,\dots; \quad 7,17,27,\dots; \quad -2,8,18,\dots \]
Why: The equation constrains only the gaps, so 3, 13, 23 and 7, 17, 27 and negative 2, 8, 18 all satisfy it. Pinning down one sequence takes one more piece of information, and the initial term is it. This is the same structure you saw with lines: a slope alone names a family, and a point picks a member out of it.
Section
Section 3
Concept
For some sequences an explicit rule is difficult to find but a recursive one is immediate. The Fibonacci sequence adds the two preceding terms; the factorials multiply by the position number.
\[ a_n = a_{n-2}+a_{n-1}; \qquad a_n = n\,a_{n-1} \]
A rule that looks back two terms needs two starting terms, one for each term the equation refers to.
Figure (svg): The Fibonacci and factorial sequences with their recursive rules
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 828-828 — Write recursive rules for special sequences
Picture it
Example 3, both parts.
Figure (svg): The Fibonacci and factorial sequences with their recursive rules
Neither has a simple explicit rule, and both have a one-line recursive one. That gap is the whole argument for recursion.
Worked example
Example 3, both parts.
\[ \text{Write recursive rules for } 1,1,2,3,5,\dots \text{ and } 1,1,2,6,24,\dots \]
First: look at how each term arises
Why: One plus 1 is 2, one plus 2 is 3, two plus 3 is 5.
First: count the starting terms needed
Why: The equation refers back two places.
\[ a _{1} = 1, a _{2} = 1 \]
Second: index from zero
Why: One is 1 times 1, two is 2 times 1, six is 3 times 2.
\[ a _{n} = n \times a _{n - 1} \]
Second: state the start
Why: The zeroth term is 1.
\[ a _{0} = 1 \]
Figure (svg): The Fibonacci and factorial sequences with their recursive rules
\[ a_1 = a_2 = 1, \; a_n = a_{n-2}+a_{n-1}; \qquad a_0 = 1, \; a_n = n\,a_{n-1} \]
Verify: extend each one step
Why: Fibonacci gives 3 plus 5, which is 8; the factorials give 5 times 24, which is 120. Both match the sequences as they continue, and the factorial values 1, 1, 2, 6, 24, 120 are the same ones you counted arrangements with in Chapter 10.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 828-828
Fill the middle
Example 3a.
Fill in the blanks
1, 1, 2, 3, 5, \; a_6 = a_4+a_5 = 3+5 = 8
Why: Three plus 5 is 8, and the sequence continues 13, 21, 34. Each term needs only the two before it, never the position number.
Worked example
Guided Practice 9.
\[ \text{Write a recursive rule for } 1,2,2,4,8,32,\dots \]
Test the usual patterns
Why: Differences 1, 0, 2, 4, 24; ratios 2, 1, 2, 2, 4.
Try combining two earlier terms
Why: One times 2 is 2, and 2 times 2 is 4.
Check further
Why: Two times 4 is 8, and 4 times 8 is 32.
State both starting terms
Why: The equation looks back two places.
\[ a _{1} = 1, a _{2} = 2 \]
Figure (svg): The solution to Worked example a two-back product shown as a ladder of expressions, one row per algebraic move
\[ a_1 = 1, \, a_2 = 2, \; a_n = a_{n-2}\,a_{n-1} \]
Verify: predict the next term and check the growth
Why: Eight times 32 is 256, so the sequence continues 1, 2, 2, 4, 8, 32, 256. Multiplying the two previous terms makes the growth faster than any geometric sequence, because the multiplier itself keeps growing.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 829-829
Trap
\[ a_1 = 1, \; a_n = a_{n-2}+a_{n-1} \]
Try to compute a sub 2
Why: The equation needs a sub zero and a sub one.
\[ a_2 = a_0+a_1 = \text{?}+1 \quad \text{(undefined)} \]
A sub zero was never given, so the sequence cannot get started. One initial term is not enough for an equation that reaches back two.
\[ a_1 = 1, \; a_2 = 1, \; a_n = a_{n-2}+a_{n-1} \]
Give as many starting terms as the equation looks back
Why: Two references back means two initial values.
\[ a_3 = a_1+a_2 = 2 \;\Longrightarrow\; 1,1,2,3,5,8,\dots \]
Count the deepest reference in the equation. That number is how many starting terms the rule requires.
Fill the middle
Example 3b.
Fill in the blanks
a_0 = 1, \; a_n = n\,a_120 \;\Longrightarrow\; a_5 = 5(24) = ___
Why: Five times 24 is 120, which is 5 factorial. This equation is unusual in mentioning n as well as the previous term.
Sorting
Count the deepest reference back.
Sort into buckets
Sort each recursive equation by how many initial terms its rule needs.
Mentioning n, as the factorial rule does, costs nothing extra: n is always available. It is references to earlier TERMS that must be paid for with initial values.
Prediction
Commit before reasoning.
Predict first
The Fibonacci sequence has the one-line recursive rule a_n = a_(n-2) + a_(n-1). What does its explicit rule look like?
Correct: Complicated — it involves powers of the golden ratio and a square root of five.
\[ a_n = \frac{\varphi^{\,n}-(1-\varphi)^{\,n}}{\sqrt{5}}, \; \varphi = \frac{1+\sqrt{5}}{2} \]
Why: An explicit rule for Fibonacci does exist, but it is built from powers of one plus root five over two divided by root five, and it produces whole numbers only by a remarkable cancellation. Nobody would find that by staring at 1, 1, 2, 3, 5. The recursive rule, by contrast, is visible in two seconds. That gap is the entire practical case for recursive rules.
Section
Section 4
Concept
A quantity that loses a fixed percentage and gains a fixed amount each period is modelled by multiplying the previous value by the retained fraction and adding the new arrivals. Such a model settles at the level where loss balances gain.
\[ a_n = 0.8a_{n-1}+5000 \]
The stable level is found by asking what value the rule leaves unchanged, which turns the recursion into a single equation.
Figure (svg): Membership numbers falling toward a stable level
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 829-829 — Use a recursive rule to model a real-life situation
Picture it
Example 4: an online music service losing 20 percent and gaining 5000 each year.
Figure (svg): Membership numbers falling toward a stable level
The yearly drops shrink and the total levels off at 25,000, where losing 20 percent costs exactly the 5000 that arrive.
Worked example
Example 4, all three steps.
\[ \text{A service has } 50{,}000 \text{ members, loses } 20\% \text{ and gains } 5000 \text{ yearly. Model it and find } a_5. \]
Find the retained fraction
Why: Losing 20 percent keeps 80 percent.
\[ 0.8 \]
Write the rule
Why: Eighty percent of last year plus 5000.
\[ a _{1} = 50, 000, a _{n} = 0.8 a _{n - 1} + 5000 \]
Step forward four years
Why: Forty-five thousand, 41,000, 37,800.
\[ a _{2}\text{ through } a _{4} \]
Compute the fifth year
Why: Eighty percent of 37,800 plus 5000.
\[ 35, 240 \]
Figure (svg): Membership numbers falling toward a stable level
\[ a_n = 0.8a_{n-1}+5000; \quad a_5 = 35{,}240 \]
Verify: watch the yearly drops
Why: They are 5000, 4000, 3200, 2560 — themselves a geometric sequence with ratio 0.8. The drops shrink because as membership falls, 20 percent of it is a smaller number, while the 5000 arriving never changes.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 829-829
Fill the middle
Example 4 Step 2.
Fill in the blanks
a_2 = 0.8(5045000000)+5000 = 40___000+5000 = ___
Why: Forty thousand retained plus 5000 new gives 45,000. The service lost 10,000 and gained 5000, for a net loss of 5000 in the first year.
Worked example
Example 4 Step 3, and Guided Practice 10.
\[ \text{What level does the membership approach? What if only } 70\% \text{ were retained?} \]
Ask what value the rule leaves alone
Why: The level stops changing when the next equals the last.
\[ x = 0.8 x + 5000 \]
Solve
Why: Zero point two x is 5000.
\[ x = 25, 000 \]
Repeat with 70 percent retained
Why: X equals 0.7x plus 5000.
\[ 0.3 x = 5000 \]
Solve
Why: Five thousand over 0.3.
\[ \text{about } 16, 667 \]
Figure (svg): The solution to Worked example the long run shown as a ladder of expressions, one row per algebraic move
\[ x = \frac{5000}{0.2} = 25{,}000; \qquad x = \frac{5000}{0.3} \approx 16{,}667 \]
Verify: check the balance at the stable level
Why: At 25,000 members the service loses 20 percent, which is 5000, and gains 5000 — exactly even. At 16,667 it loses 30 percent, which is 5000, and gains 5000. In both cases the stable level is the new arrivals divided by the loss rate, which is worth remembering as a shortcut.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 829-829
Trap
\[ \text{loses } 20\% \text{ each year} \]
Multiply the previous year by 0.2
Why: The percentage lost is used as the multiplier.
\[ a_2 = 0.2(50{,}000)+5000 = 15{,}000 \quad \text{(wrong)} \]
Losing a fifth means keeping four fifths. The multiplier is what REMAINS, not what leaves.
\[ a_n = 0.8a_{n-1}+5000 \]
Multiply by the retained fraction
Why: One minus the loss rate.
\[ a_2 = 0.8(50{,}000)+5000 = 45{,}000 \]
This is the same 1 minus p that turned percent decay into a ratio in Lesson 12.3, reappearing inside a recursive rule.
Fill the middle
Example 4 Step 3.
Fill in the blanks
x = 0.8x+5000 \;\Longrightarrow\; 0.2x = 5000 \;\Longrightarrow\; x = 25000
Why: Five thousand over 0.2 is 25,000. At that level 20 percent lost is exactly the 5000 gained, so nothing changes from year to year.
Ranking
Smallest first.
Put in order
Why: The values are about 16,667, then 25,000, 35,240, 45,000 and 50,000. Retaining less drives the stable level down sharply: a ten-point drop in retention costs a third of the eventual membership.
Prediction
Commit before reasoning.
Predict first
The service loses 20 percent of its members every year forever. Why does the membership not fall to zero?
Correct: Because 5000 new members arrive each year, and as membership falls the 20 percent lost becomes smaller than 5000.
\[ 0.2x = 5000 \;\Longleftrightarrow\; x = 25{,}000 \]
Why: At 50,000 members the loss is 10,000 against a gain of 5000, a net fall. At 20,000 members the loss is only 4000 against the same gain of 5000, a net RISE. Somewhere between the two the forces balance, and that crossing point is the stable level of 25,000. A model with a fixed inflow and a proportional outflow always has one, which is why this shape appears in so many population and drug-dosage problems.
Section
Section 5
Concept
Iteration is the repeated composition of a function with itself. Starting from an initial value, each output becomes the next input, which generates a sequence recursively.
\[ x_1 = f(x_0), \; x_2 = f(x_1), \; x_3 = f(x_2), \dots \]
An iterate is simply a number produced by this process. Iterating f is exactly the recursive rule a_n equals f of a_(n-1).
Figure (svg): A function iterated three times from a starting value
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 830-830 — Iterating functions
Picture it
Example 5: iterating negative 3x plus 1 from x equal to 2.
Figure (svg): A function iterated three times from a starting value
Three applications take 2 to negative 5 to 16 to negative 47. The negative coefficient makes the signs alternate and the sizes grow.
Worked example
Example 5.
\[ \text{Find } x_1, x_2, x_3 \text{ for } f(x) = -3x+1 \text{ with } x_0 = 2. \]
Compute the first iterate
Why: F of 2 is negative 6 plus 1.
\[ x _{1} = -5 \]
Feed that back in
Why: F of negative 5 is 15 plus 1.
\[ x _{2} = 16 \]
Feed that back in
Why: F of 16 is negative 48 plus 1.
\[ x _{3} = -47 \]
Note the behaviour
Why: Signs alternate, sizes grow.
Figure (svg): A function iterated three times from a starting value
\[ x_1 = -5, \; x_2 = 16, \; x_3 = -47 \]
Verify: check what the multiplier is doing
Why: Each step multiplies by negative 3 and adds 1, so the size roughly triples every time and the sign flips. This is the same multiply-then-add shape as the music service model, but with a multiplier larger than 1 in size the iterates run away instead of settling.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 830-830
Fill the middle
Example 5.
Fill in the blanks
x_2 = f(x_1) = f(-5) = -3(-5)+1 = 16
Why: Negative 3 times negative 5 is 15, plus 1 gives 16. The input was the previous OUTPUT, not the original starting value.
Worked example
Guided Practice 11 and 12.
\[ \text{Iterate } f(x) = 4x-3 \text{ from } x_0 = 2, \text{ and } f(x) = x^2-5 \text{ from } x_0 = -1. \]
First: f of 2
Why: Eight minus 3.
\[ x _{1} = 5 \]
First: continue
Why: F of 5 is 17, and f of 17 is 65.
\[ 5, 17, 65 \]
Second: f of negative 1
Why: One minus 5.
\[ x _{1} = -4 \]
Second: continue
Why: F of negative 4 is 11, and f of 11 is 116.
\[ -4, 11, 116 \]
Figure (svg): The solution to Worked example two more functions shown as a ladder of expressions, one row per algebraic move
\[ 5,17,65; \qquad -4,11,116 \]
Verify: compare the two growth patterns
Why: The linear function roughly quadruples each step, while the quadratic one squares — going from 11 to 116 in a single application. Iterating a quadratic accelerates far faster than iterating a line, which is why the second sequence overtakes the first by the third iterate.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 830-830
Error analysis
A student iterates f(x) = -3x + 1 from x_0 = 2 and reports the second iterate.
Annotate
On: \( x_2 = f(x_0) \text{ twice} = -3(2)+1 = -5 \)
Writing the chain out as 2 to -5 to 16 to -47, with an arrow for each application, makes it impossible to lose your place.
Matching
Feed each output back in.
Match the pairs
Why: The last is Example 4 restated: a recursive model IS an iteration, with the model's rule as the function. The first three run away; the fourth settles, because its multiplier is under 1 in size.
Comparison
Fill the blanks. Both iterate multiply-then-add.
Comparison matrix
| Question | f(x) = -3x + 1 | f(x) = 0.8x + 5000 |
|---|---|---|
| Size of the multiplier | 3 | 0.8 |
| Behaviour | runs away | settles at a level |
| Fixed value | 0.25, but never approached | 25,000, and approached |
| Signs | alternate | stay positive |
Both functions have a value they leave unchanged, but only the one with multiplier under 1 in size actually pulls the iterates toward it. That threshold at 1 is the same one that decided whether an infinite series had a sum.
Prediction
Commit before reasoning.
Predict first
What is the connection between iterating a function f and a recursive rule?
Correct: Iterating f is exactly the recursive rule a_n = f(a_(n-1)), with x_0 as the initial term.
\[ x_n = f(x_{n-1}) \;\Longleftrightarrow\; a_n = f(a_{n-1}) \]
Why: Every recursive equation in this lesson can be read as a function applied to the previous term: adding 4 is the function x plus 4, tripling is 3x, and the music service's rule is 0.8x plus 5000. Iteration is the same idea presented as repeated composition rather than as a sequence, which is why the two sections sit in one lesson. Where the equation mentions n as well, as the factorial rule does, the function changes at each step and the framing is less natural.
Comparison
Fill the blanks. Same sequences, different descriptions.
Comparison matrix
| Question | Explicit | Recursive |
|---|---|---|
| Gives a_n in terms of | the position n | the preceding term or terms |
| Arithmetic form | a_n = a_1 + (n - 1)d | a_n = a_(n-1) + d |
| Geometric form | a_n = a_1 times r to the n - 1 | a_n = r a_(n-1) |
| Best for | reaching a distant term | describing a step-by-step process |
Neither form is better in general. Explicit rules answer what is the hundredth term; recursive rules answer what happens next, and real processes usually present themselves in the second way.
Pattern
Start, step, then run.
Supply as many initial terms as the equation reaches back. One reference back needs one; two need two.
OpenStax Algebra and Trigonometry 2e, §13.1 Sequences and Their Notations §13.1
Check
Start at the given term and step forward.
Check your understanding
What are the first four terms for a_1 = 1, a_2 = 4, and a_n = a_(n-1) times a_(n-2)?
Answer: A
Why: a_3 = 4 times 1 = 4, and a_4 = 4 times 4 = 16.
Check
Both halves of the rule are needed.
Check your understanding
What is a complete recursive rule for 16, 40, 100, 250, ...?
Answer: A
Why: The ratio is 40/16 = 2.5, and the rule must also state the first term.
Check
Each output becomes the next input.
Check your understanding
What are the first three iterates of f(x) = 4x - 3 for x_0 = 2?
Answer: A
Why: f(2) = 5, then f(5) = 17, then f(17) = 65.
Real world
A patient takes a 300 mg dose of a drug every 8 hours. In each 8-hour period the body eliminates 60 percent of whatever is present.
Discussion prompt
Write a recursive rule for the amount just after the nth dose, and find the level it settles toward.
Hint: What fraction remains from the previous dose when the next one arrives?
Answer:
\[ a_1 = 300, \quad a_n = 0.4a_{n-1}+300 \]
\[ a_2 = 420, \; a_3 = 468, \; a_4 = 487.2, \; a_5 = 494.9 \]
\[ x = 0.4x+300 \;\Longrightarrow\; 0.6x = 300 \;\Longrightarrow\; x = 500 \]
The amount in the body settles toward 500 mg, reached from below and never exceeded.
This is the music service model wearing a lab coat, and it is why dosing schedules are designed the way they are: the steady-state level is the dose divided by the fraction eliminated, so halving the elimination rate doubles the level the patient ends up carrying. A drug whose elimination is slow enough can reach a dangerous steady state from doses that each look perfectly safe on their own — the recursion, not the single dose, is what determines the risk.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is a_n = a_(n-1) + 10 a complete recursive rule for the sequence 3, 13, 23, 33?
Correct: No — without an initial term it also describes 7, 17, 27 and infinitely many others.
\[ a_1 = 3, \; a_n = a_{n-1}+10 \]
Why: The equation constrains only the gaps between terms, never where the sequence begins. Adding a_1 equal to 3 pins down exactly one sequence. The textbook flags this on page 828 as the error to avoid, and it is the same logical structure as a line: a slope names a whole family of parallel lines, and one point selects a single member.
Explain it
They know explicit rules and have never seen a recursive one.
Discussion prompt
In four sentences or fewer, explain what a recursive rule is and why anyone would use one.
Hint: Compare a recipe with a lookup table.
Answer:
An explicit rule tells you a term straight from its position, like looking up an answer in a table. A recursive rule instead tells you where to start and what to do to get from each term to the next.
It is slower for reaching a distant term, because you have to walk the whole way. But some sequences are easy to describe step by step and very hard to describe by position, and Fibonacci is the standard example.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the initial term, count how far back the equation reaches and supply that many. For the notation, read a sub n minus 1 aloud as the term before this one. For models, multiply by what REMAINS and add what arrives. For the stable level, set x equal to the rule applied to x and solve the resulting equation.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a recursion page. Top left: take one sequence and write it BOTH ways, explicit and recursive, then compute its hundredth term by whichever rule is faster and note beside it how many steps the other would have taken. Top right: write the two standard recursive equations for arithmetic and geometric sequences, and beside them write one incomplete rule and one complete rule, marking what is missing from the first. Middle: write the Fibonacci and factorial rules, generate six terms of each, and note how many initial terms each one needs and why. Bottom left: model a repeating real-world process of your own invention, generate five terms, and solve for the level it settles toward. Bottom right: iterate a function of your choice three times, drawing the chain of arrows, and say whether the iterates run away or settle.
If any recursive rule on your page lacks a starting term, it names infinitely many sequences rather than one. Add the start.
Recap
Five things, and a second language for describing a sequence.
| If you see | Then |
|---|---|
| a_(n-1) in a rule | Read it as the previous TERM, not as n minus 1 |
| A constant added each step | The sequence is arithmetic; d is that constant |
| A constant multiplier each step | The sequence is geometric; r is that multiplier |
| An equation reaching back two terms | Supply two initial terms |
| Loses p percent, gains a fixed amount | Multiply by 1 minus p, then add the amount |
| A level that stabilises | Solve x equals the rule applied to x |
That closes Chapter 12. Chapter 13 leaves sequences behind and builds trigonometry from right triangles, starting with the six ratios and ending with the laws of sines and cosines.
McDougal Littell Algebra 2 (Texas Edition), Ch. 12 Sequences and Series — Lesson 12.5 Use Recursive Rules with Sequences and Functions §12.5, pp. 827-831 — everything on these slides traces back here
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