3.1 - Generating and describing sequences

3.1 - Generating and describing sequences

Learn to build a sequence one term at a time or jump straight to a chosen position. You will compare familiar integer patterns and describe a rule clearly enough for someone else to continue it.

Term-to-term rules

A sequence starts at 44 and repeatedly doubles, then subtracts 11. Will its fourth term be 2525 or 3131? The order of the operations decides the answer.

You already know

Brackets keep a calculation together. Multiplication comes before subtraction unless brackets change that order.

Start with the actual number 44: doubling gives 88, then subtracting 11 gives 77. For the next step, use 77, not the starting number again.

Sequence

A sequence is an ordered list of numbers called terms. A term-to-term rule tells you how to obtain the next term from earlier terms. For example, “start at 22, then add 33” gives 2,5,8,11,…2, 5, 8, 11, \ldots.

A starting term matters. “Add 33” alone could describe many sequences. Neither the Foundation nor Higher formulae sheet supplies sequence rules: use the rule given in the question, or describe the pattern you find.

Worked example 1: follow a compound rule

Question. The first term is 44. Each next term is twice the previous term, minus 11. Write the first four terms.

Plan. Start with 44, then repeat the two operations on each new term.

2(4)−1=7second term2(7)−1=13third term2(13)−1=25fourth term\begin{aligned} 2(4)-1&=7 &&\text{second term}\\ 2(7)-1&=13 &&\text{third term}\\ 2(13)-1&=25 &&\text{fourth term} \end{aligned}

Why use 1313 in the last line? The rule acts on the previous term's value.

Answer. 4,7,13,254, 7, 13, 25.

Check. Undo the last step: (25+1)÷2=13(25+1)\div2=13.

Now change only the starting term to 11. The same operations can produce a very different pattern: a sequence need not keep increasing.

Worked example 2: a term that stays fixed

Question. Start at 11. Double the previous term, then subtract 11. Write the first four terms.

Plan. Apply exactly the same rule at every step.

2(1)−1=1second term2(1)−1=1third term2(1)−1=1fourth term\begin{aligned} 2(1)-1&=1 &&\text{second term}\\ 2(1)-1&=1 &&\text{third term}\\ 2(1)-1&=1 &&\text{fourth term} \end{aligned}

Why does it stay at 11? The result of the rule is the same as the input, so the next step starts in the same place.

Answer. 1,1,1,11, 1, 1, 1.

Check. Another application gives 2(1)−1=12(1)-1=1, so the pattern continues.

Keep the words “then subtract” attached to the result of doubling.

Watch out

The wrong line 2(4−1)=62(4-1)=6 subtracts first. “Double, then subtract 11” means 2(4)−1=72(4)-1=7. Brackets can change the rule.

Your turn changes only the starting term again.

Position-to-term rules

A term-to-term rule is like walking along a line of stepping stones. A position-to-term rule lets you choose a numbered stone straight away. The position nn starts at 11, then takes positive integer values.

Position-to-term rule

A position-to-term rule calculates a term directly from its position. The notation unu_n means “the term in position nn”. The subscript is a label: u4u_4 is the fourth term, not uu multiplied by 44.

For instance, un=2nu_n=2n gives the even numbers 2,4,6,…2, 4, 6, \ldots, while un=2n−1u_n=2n-1 gives the odd numbers 1,3,5,…1, 3, 5, \ldots. Substitute the position into the whole rule, then follow the usual order of operations.

Worked example 3: jump to a distant term

Question. A sequence has rule un=n2−2u_n=n^2-2. Find its first four terms and its tenth term.

Plan. Replace nn by each requested position and square before subtracting.

u1=12−2=−1position oneu2=22−2=2position twou3=32−2=7position threeu4=42−2=14position fouru10=102−2=98position ten\begin{aligned} u_1&=1^2-2=-1 &&\text{position one}\\ u_2&=2^2-2=2 &&\text{position two}\\ u_3&=3^2-2=7 &&\text{position three}\\ u_4&=4^2-2=14 &&\text{position four}\\ u_{10}&=10^2-2=98 &&\text{position ten} \end{aligned}

Why substitute 1010, rather than the ninth term? This rule uses the position directly.

Answer. −1,2,7,14-1, 2, 7, 14; the tenth term is 9898.

Check. Adding 22 back gives 1,4,9,161, 4, 9, 16, the expected squares.

Change only the operation on the position from squaring to cubing. A small change to the rule makes the terms grow much faster.

Worked example 4: use powers of the position

Question. A sequence has rule vn=n3−2v_n=n^3-2. Find its first four terms and its tenth term.

Plan. Cube each position, then subtract 22.

v1=13−2=−1position onev2=23−2=6position twov3=33−2=25position threev4=43−2=62position fourv10=103−2=998position ten\begin{aligned} v_1&=1^3-2=-1 &&\text{position one}\\ v_2&=2^3-2=6 &&\text{position two}\\ v_3&=3^3-2=25 &&\text{position three}\\ v_4&=4^3-2=62 &&\text{position four}\\ v_{10}&=10^3-2=998 &&\text{position ten} \end{aligned}

Why calculate the power first? The subtraction is outside the power in the stated rule.

Answer. −1,6,25,62-1, 6, 25, 62; the tenth term is 998998.

Check. The fourth result gives 62+2=64=4362+2=64=4^3.

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