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Two kinds of rule

Patterns and sequences · level 4, lesson 1 of 6

A question to start

Two friends describe the same list of numbers, 5, 8, 11, 14, 17, ...

  • Ana says: "Start at 5. Then keep adding 3."
  • Ben says: "Take the position, multiply it by 3, and add 2."

They both work. Check Ben's on the 4th number: \(4 \times 3 + 2 = 14\). ✓ But they are clearly different kinds of rule. Ask each of them for the 1000th number. Ana needs 999 additions. Ben needs one multiplication and one addition: 3002. And ask each of them, "Is 400 in the list?" Who can answer faster, and how?

The idea

Names for the parts

To talk about any position, not just the 4th or the 10th, we use a letter. The usual one is \(n\), for "number of the position". The term in position \(n\) is written \(t_n\), read "t sub n" or "t n". The small \(n\) is the position; \(t_n\) is the term that lives there.

\(n\) (position) 1 2 3 4 5
\(t_n\) (term) 5 8 11 14 17

So \(t_1 = 5\), \(t_3 = 11\), \(t_5 = 17\). If we write \(t_{10}\) we mean "the 10th term", whatever it turns out to be.

Rule 1: the next-term rule (recursive)

Ana's rule tells you how to get from one term to the next. Written with symbols:

\[t_1 = 5, \qquad t_{n+1} = t_n + 3\]

Read the second part as: "the term after \(t_n\) is \(t_n\) plus 3". A rule that makes each term out of earlier terms is called recursive (from a Latin word for "running back"), because to use it you keep running back to the term before.

A recursive rule needs two pieces: a starting value and a step. "Add 3" on its own does not decide the sequence: 5, 8, 11, ... and 1, 4, 7, ... and 100, 103, 106, ... all follow it. The start pins down which one we mean.

Rule 2: the position rule (explicit)

Ben's rule tells you the term straight from its position:

\[t_n = 3n + 2\]

\(3n\) means \(3 \times n\). Put in a position, get out the term. A rule like this is called explicit, because it spells out the answer directly, with no running back.

5 +3 8 +3 11 +3 14 ... next-term rule: start at 5, then add 3 each time n × 3 + 2 3n + 2 position rule: put in the position n, get out the term

Why the two rules agree

Are we sure Ben's rule gives the same numbers as Ana's for every position, not just the five we checked? Use the jumps from level 3. To reach position \(n\), Ana starts at 5 and makes \(n - 1\) jumps of 3:

\[t_n = 5 + 3(n - 1) = 5 + 3n - 3 = 3n + 2\]

(Here \(3(n - 1)\) means "3 lots of \(n - 1\)", which is 3 lots of \(n\) take away 3 lots of 1: \(3n - 3\).) So the two rules are the same sequence, described two ways. That is not a check of a few terms; it is a reason that covers every \(n\) at once.

Which rule is better?

Next-term (recursive) Position (explicit)
Easy to spot from a list? Yes: just look at the gaps Harder: you have to work it out
Find the 1000th term 999 steps One calculation
Is 400 in the list? Keep going until you pass 400 Solve \(3n + 2 = 400\)
Works for every pattern? Very often Sometimes very hard to find

Each has its job. In real life, a spreadsheet uses recursive rules all the time ("this cell = the cell above + 3"), because computers do not mind doing 999 steps. People prefer explicit rules, because we do mind.

Is a number in the sequence?

With an explicit rule, "Is 400 a term?" becomes "Is there a whole number \(n\) with \(3n + 2 = 400\)?" Take away 2: \(3n = 398\). Then \(n = 398 \div 3 = 132.66...\), not a whole number. So 400 is not a term: it falls between the 132nd term (\(3 \times 132 + 2 = 398\)) and the 133rd (401). Positions must be whole numbers: there is no "132.66th term".

Other kinds of sequence have both rules too

Sequence Recursive rule Explicit rule
Square numbers 1, 4, 9, 16, ... \(t_1 = 1\); add the next odd number \(t_n = n^2\)
Doubling 1, 2, 4, 8, ... \(t_1 = 1\), \(t_{n+1} = 2t_n\) \(t_n = 2^{n-1}\)
Fibonacci 1, 1, 2, 3, 5, ... \(t_1 = t_2 = 1\), \(t_{n+2} = t_{n+1} + t_n\) exists, but needs level 7 and 8 ideas

\(2^{n-1}\) means \(n - 1\) twos multiplied together, the same "one less than the position" as the jumps: the 1st term has had no doublings yet. The Fibonacci sequence shows that a recursive rule can be very simple while the explicit one is very hard.

Worked examples

1. Both rules for 7, 11, 15, 19, ...

  • Recursive: the gap is always 4, so \(t_1 = 7\), \(t_{n+1} = t_n + 4\).
  • Explicit: \(n - 1\) jumps of 4 from 7: \(t_n = 7 + 4(n - 1) = 7 + 4n - 4 = 4n + 3\).
  • Check: \(t_2 = 4 \times 2 + 3 = 11\) ✓, \(t_4 = 19\) ✓.
  • The 50th term: \(t_{50} = 4 \times 50 + 3 = 203\).

2. From an explicit rule to a list. \(t_n = 5n - 1\).

  • \(t_1 = 5 - 1 = 4\), \(t_2 = 10 - 1 = 9\), \(t_3 = 14\), \(t_4 = 19\).
  • The gap is 5 every time, so the recursive rule is \(t_1 = 4\), \(t_{n+1} = t_n + 5\).

3. From a recursive rule to a list. \(t_1 = 2\), \(t_{n+1} = 2t_n + 1\) ("double it and add 1").

  • \(t_2 = 2 \times 2 + 1 = 5\), \(t_3 = 2 \times 5 + 1 = 11\), \(t_4 = 23\), \(t_5 = 47\).
  • The gaps are 3, 6, 12, 24: not constant, so the explicit rule will not look like "a number times \(n\), plus a number". (It is \(t_n = 3 \times 2^{n-1} - 1\); check \(t_5 = 3 \times 16 - 1 = 47\) ✓. You will learn to find rules like this in level 6.)

4. Is it a term? For \(t_n = 3n + 2\), is 200 a term? Is 250?

  • \(3n + 2 = 200\) gives \(3n = 198\), \(n = 66\). A whole number: yes, 200 is the 66th term.
  • \(3n + 2 = 250\) gives \(3n = 248\), \(n = 82.66...\). No.

Try it

  1. Write a recursive rule and an explicit rule for 6, 10, 14, 18, ...
  2. Write the first five terms of \(t_n = 2n + 7\).
  3. Write the first five terms of \(t_1 = 3\), \(t_{n+1} = 2t_n\).
  4. For \(t_n = 6n - 4\), find \(t_{20}\). Is 100 a term?
  5. Write the first five terms of \(t_n = n^2 + 1\). Describe the gaps between them.
  6. The rule "\(t_{n+1} = t_n + 3\)" is given with no starting value. Write two different sequences that follow it.
  7. Cara says "start at 5 and add 3" is the same as \(t_n = 3n + 5\). Is she right? Find the correct explicit rule.
  8. A sequence has \(t_1 = 100\) and \(t_{n+1} = t_n - 7\). Find an explicit rule, and the first negative term.
Answers
  1. Recursive: \(t_1 = 6\), \(t_{n+1} = t_n + 4\). Explicit: \(t_n = 6 + 4(n - 1) = 4n + 2\). (Check: \(t_3 = 14\) ✓.)
  2. 9, 11, 13, 15, 17.
  3. 3, 6, 12, 24, 48.
  4. \(t_{20} = 6 \times 20 - 4 = 116\). For 100: \(6n - 4 = 100\) gives \(6n = 104\), \(n = 17.33...\), so no.
  5. 2, 5, 10, 17, 26. The gaps are 3, 5, 7, 9: the odd numbers, just as for the square numbers (each term is one more than a square).
  6. For example 1, 4, 7, 10, ... and 10, 13, 16, 19, ... Any start works, which is why a recursive rule needs a starting value.
  7. No. \(3n + 5\) gives \(t_1 = 8\), not 5. Five is the start, and the start has had no 3s added. The correct rule is \(t_n = 5 + 3(n - 1) = 3n + 2\).
  8. \(t_n = 100 - 7(n - 1) = 107 - 7n\). It is negative when \(7n > 107\), that is \(n > 15.28...\), so the first negative term is \(t_{16} = 107 - 112 = -5\). (Just before it, \(t_{15} = 107 - 105 = 2\).)

Common mistake

Writing "start at \(a\), add \(d\)" as \(dn + a\). The start is the 1st term, and it has had no \(d\) added yet. So the rule is \(a + d(n - 1)\), which is \(dn + (a - d)\). Always test your explicit rule with \(n = 1\): it must give the first term.

Mixing up \(n\) and \(t_n\). In "is 200 a term?", 200 is a value of \(t_n\), so you solve for \(n\). Putting \(n = 200\) into the rule answers a different question (what is the 200th term?).

Where it leads