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Stage 1 - Stage 3

Lesson

An ordered list of numbers, separated by commas, is called a *progression* or *sequence*. For example $-3,5,13,21...$−3,5,13,21... and $1,10,100,1000...$1,10,100,1000... are two interesting mathematical progressions. If the sequence ends, it is known as a finite sequence. Otherwise it is said to be infinite. You might notice that in both examples there is a pattern in the way the progressions are formed. In the first progression, each term, other than the first, is eight more than the previous term in the list. In the second progression, each term, apart from the first, is ten times more than the previous term. When a pattern is detectable in a progression, a generating rule can often be established that allows us to determine any term in the sequence.

Sequences can have strange generating rules like this one provided by the mathematician John Conway. It begins $1,11,21,1211,111221,\dots$1,11,21,1211,111221,… and so on. It starts with the number $1$1, and the next term is created by observing that the first term is really “one $1$1”. The third term is created by observing that the second term is really “two 1s”. The fourth term likewise is generated as “one $2$2, and one $1$1”. The process of generation continues indefinitely.

There are some sequences that appear to have no pattern at all, but nevertheless have a certain logical way of building. For example, the sequence $3,1,4,1,5,9,\dots$3,1,4,1,5,9,… separating the digits of $\pi$π exhibits no discernable pattern and continues to do so indefinitely. There are strange sequences such as $7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1\dots$7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1,4,2,1,4,2,1… which begin with any positive integer (here, 7 was chosen) and then, depending on the parity of the starting term, one of two rules are used to generate the next term. If the term is odd, triple it and add $1$1. If it’s even, halve it. It seems the sequence always ends with the three numbers $4$4, $2$2 and $1$1, no matter what the starting number.

Mathematicians can sometimes develop explicit generating rules that allow the calculation of any particular term in the sequence. For example the rule $a_n=\sqrt{n}$`a``n`=√`n` means that the $n$`n`th term is the square root of n. So the first term becomes $a_1=\sqrt{1}=1$`a`1=√1=1 and the second term $a_2=\sqrt{2}$`a`2=√2 etc., so that the sequence becomes $1,\sqrt{2},\sqrt{3},2,...$1,√2,√3,2,...and so on. Sometimes these explicit rules can be a little hard to find because the pattern exhibited by the first few terms may be obscure. For example, the pattern for the sequence beginning $\log8,\log64,\log512,...$`l``o``g`8,`l``o``g`64,`l``o``g`512,... may not be immediately obvious, however by using the rules of logarithms, we can change the look of the terms to $\log8,2\times\log8,3\times\log8,...$`l``o``g`8,2×`l``o``g`8,3×`l``o``g`8,...

We can express a sequence using a recurrence equation where each new term is generated by some function of a previous term or terms. Take for example, the sequence described by:

$a_n=2a_{n-1}+n,a_1=3$`a``n`=2`a``n`−1+`n`,`a`1=3

This expression tells us that the first term $a_1$`a`1 is $3$3. Then it shows us how to find, progressively, any other term in the sequence. For example, looking at the formula, the second term $a_2$`a`2 is equal to twice the first term $a_1$`a`1 plus $2$2, which is $2\times3+2$2×3+2 or $8$8. The third term is equal to twice the second term plus $3$3, or in other words $19$19. The fourth term is twice $19$19 plus $4$4, or $42$42. The fifth term is twice $42$42 plus $5$5, or $89$89. This process of deducing the $n$`n`th term from the $\left(n-1\right)$(`n`−1)th term can continue indefinitely. A simple computer spread sheet program could develop a list of terms very easily.

Is the sequence $1,2,3,4,5,6$1,2,3,4,5,6 finite or infinite?

Finite

AInfinite

B

State the first five terms of the sequence $a_n=3n-3$`a``n`=3`n`−3, starting from $n=1$`n`=1.

Write all five terms on the same line separated by a comma.

Consider the sequence defined by $a_1=3$`a`1=3, $a_n=a_{n-1}+n$`a``n`=`a``n`−1+`n` for $n>1$`n`>1.

State the first term.

State the second term.

State the third term.

State the fourth term.