6.06a, 6.06b - Rules for sequences

6.06a, 6.06b - Rules for sequences

How could you find the 100th number in a pattern without listing the first 99? Learn the difference between a term and its position, generate sequences from rules, find arithmetic nth terms and recognise triangular, square and cube numbers.

Term-to-term rules

Suppose a row of seats contains 5 chairs, the next row 8, then 11, then 14. To continue this seating pattern, add 3 to the previous number each time.

Sequence

A sequence is an ordered list of numbers called terms. A term-to-term rule tells you how to obtain a term from the preceding term or terms. A starting value is needed as well as the rule.

The position is the place in the list, counting from 1 unless told otherwise. In 5,8,11,14,…5,8,11,14,\ldots, the third term is 11: 3 is its position, not its value. The dots mean the list continues according to its rule.

An arithmetic sequence adds the same number each time. This signed number is the common difference. It can be positive, negative or zero. A rule may also involve multiplication or several operations; not every sequence is arithmetic.

Worked example 1: an increasing sequence

Question. Start at 5 and add 3 each time. Write the first five terms.

Plan. Keep the starting term, then apply the rule four times.

5+3=88+3=1111+3=1414+3=17\begin{aligned} 5+3&=8 \\ 8+3&=11 \\ 11+3&=14 \\ 14+3&=17 \end{aligned}

Answer. 5,8,11,14,175,8,11,14,17.

Check. Subtracting consecutive terms gives 8−5=11−8=14−11=17−14=38-5=11-8=14-11=17-14=3. There are five terms but only four moves.

Change only the sign of the common difference. Adding a negative difference means subtracting.

Worked example 2: a decreasing sequence

Question. Start at 5 and add −3-3 each time. Write the first five terms.

Plan. Subtract 3 at each move, continuing past zero.

5−3=22−3=−1−1−3=−4−4−3=−7\begin{aligned} 5-3&=2 \\ 2-3&=-1 \\ -1-3&=-4 \\ -4-3&=-7 \end{aligned}

Answer. 5,2,−1,−4,−75,2,-1,-4,-7.

Check. Reversing each move adds 3; for example, −7+3=−4-7+3=-4. The signed difference is consistently −3-3.

A rule can use more than one operation. Starting at 4 with “multiply by 2, then subtract 1” gives 2×4−1=72\times4-1=7, then 2×7−1=132\times7-1=13. Apply the whole rule each time.

A finite list suggests a rule but does not prove that rule must continue. For example, both “keep adding 3” and “add 3 for four moves, then add 10” start 5,8,11,14,175,8,11,14,17. Use the rule or sequence class stated in the question; when spotting a pattern, say which rule you are using.

Position-to-term rules

A position-to-term rule uses the position itself, so you can jump straight to a distant term.

The sequence 2,4,6,8,…2,4,6,8,\ldots, defined as the positive even numbers, has the rule “double the position”: its nth term is 2n2n. The sequence 3,4,5,6,…3,4,5,6,\ldots, defined by adding 1, has nth term n+2n+2. Here nn is a positive integer: 1, 2, 3 and so on. A position of 2.52.5 does not name a term in these sequences.

Nth term

An nth-term formula is a position-to-term rule. Substitute the position for nn to obtain the term's value. The expression 3n+23n+2 means multiply the position by 3, then add 2; it does not mean add 3n+23n+2 to the previous term.

Use the order of operations whenever you substitute. A supplied formula can contain powers even when the sequence is not arithmetic.

Worked example 3: using a linear rule

Question. A sequence has nth term 3n+23n+2. Find its first three terms and its 20th term.

Plan. Substitute n=1,2,3n=1,2,3 and 20 separately.

n=1:3(1)+2=5n=2:3(2)+2=8n=3:3(3)+2=11n=20:3(20)+2=62\begin{aligned} n=1:&\quad3(1)+2=5 \\ n=2:&\quad3(2)+2=8 \\ n=3:&\quad3(3)+2=11 \\ n=20:&\quad3(20)+2=62 \end{aligned}

Answer. The first three terms are 5,8,115,8,11; the 20th term is 62.

Check. The common difference is 3. From the first to the 20th term there are 19 moves, so 5+19×3=625+19\times3=62.

Now change the formula to include a square. Substitute the position in every place that nn appears.

Worked example 4: using a rule with a power

Question. A sequence has nth term n2+2nn^2+2n. Find its first three terms and its 20th term.

Plan. Square the position, then add twice that same position.

n=1:12+2(1)=3n=2:22+2(2)=8n=3:32+2(3)=15n=20:202+2(20)=440\begin{aligned} n=1:&\quad1^2+2(1)=3 \\ n=2:&\quad2^2+2(2)=8 \\ n=3:&\quad3^2+2(3)=15 \\ n=20:&\quad20^2+2(20)=440 \end{aligned}

Answer. 3,8,153,8,15; the 20th term is 440.

Check. Rewrite n2+2n=n(n+2)n^2+2n=n(n+2). At position 20 this gives 20×22=44020\times22=440, agreeing with the substitution.

The differences 5,7,…5,7,\ldots in the second example are not constant. An nth-term rule need not describe an arithmetic sequence.

Finding an arithmetic nth term

If the common difference is constant, count how many equal moves take you from the first term to position nn.

There are no moves to reach the first term, one to reach the second, and n−1n-1 to reach the nth. That explains the rule below, including its easily missed minus 1.

Continue this lesson.

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