Functions · Book or PDF

Functions 11

Chris Kirkpatrick, Barbara Alldred, Andrew Dmytriw, Shawn Godin, Angelo Lillo, David Pilmer, Susanne Trew, Noel Walker · Grade 11 · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 7 · pp. 422–481

Chapter 7: Discrete Functions: Sequences and Series

This chapter studies lists of numbers that follow a rule. It starts with sequences in general, then looks closely at arithmetic sequences (constant added amount) and geometric sequences (constant multiplier). It then explores other sequences such as the Fibonacci sequence and the golden ratio. Finally, it adds up the terms of sequences to form series and uses Pascal's triangle to expand binomials.

Sequences, Terms and Recursive Rules

A sequence is an ordered list of numbers. Order matters because each number is tied to a position. The lists 4, 12, 20, 28, ... and 12, 4, 20, 28, ... use the same numbers but are different sequences (p. 426).

Each number in the list is a term. A subscript shows its position, so $t_1$ is the 1st term and $t_{12}$ is the 12th. The general term $t_n$ is a formula that gives any term from its position $n$ . If $t_n = 2n$ , then $t_{12} = 2(12) = 24$ (p. 426).

A sequence can also be described recursively. You give the starting term (or terms) and a recursive formula that builds each new term from the earlier one(s). For example, $t_1 = 1$ , $t_n = t_{n-1} + 2$ gives 1, 3, 5, 7, ... You must know the previous term before you can find the next (pp. 426, 428).

Sequences as Discrete Functions

Every position number $n$ gives exactly one term, so every sequence is a function. Its domain is the natural numbers $\{1, 2, 3, \dots\}$ . Its range is the set of terms. For 4, 12, 20, 28, ... the domain is $\{1, 2, 3, 4, \dots\}$ and the range is $\{4, 12, 20, 28, \dots\}$ . The function is discrete, so its graph is separate points, not a continuous curve (p. 433).

Arithmetic Sequences

An arithmetic sequence has the same difference between any two consecutive terms. This constant is the common difference, $d$ . In 2, 6, 10, 14, ... each term is 4 more than the one before, so $d = 4$ . In 9, 6, 3, 0, ... we get $d = -3$ , so the sequence decreases (pp. 426, 433). To test a sequence, subtract each term from the next and see whether the answers are all equal.

General term. To reach the $n$ th term, start at the first term $a$ and add $d$ a total of $n - 1$ times:

$t_n = a + (n-1)d$

For 3, 12, 21, 30, ... we have $a = 3$ and $d = 9$ . Then $t_n = 3 + (n-1)(9) = 9n - 6$ . One way to see this: the multiples of 9 are 9, 18, 27, ..., and each term of our sequence is 6 less than the multiple of 9 in the same position (pp. 427, 433).

Recursive formula. Each new term is the previous term plus $d$ :

$t_1 = a,\quad t_n = t_{n-1} + d \quad (n > 1,\ n \in N)$

For 3, 12, 21, ... this is $t_1 = 3$ , $t_n = t_{n-1} + 9$ . It is easy for finding the next term but slow for a term far down the list (pp. 428, 433).

Linear function. Constant first differences mean the points (position, term) lie on a straight line. So an arithmetic sequence is a discrete linear function

$f(n) = dn + b,\quad b = t_0 = a - d$

The slope is $d$ . The intercept $t_0$ is not a term, because $n$ starts at 1. For 18, 11, 4, $-3$ , ... the slope is $-7$ and $b = 18 - (-7) = 25$ , so $f(n) = -7n + 25$ (pp. 429, 433).

Finding Terms of an Arithmetic Sequence

Finding a specific term. Identify $a$ and $d$ , then substitute them and $n$ into $t_n = a + (n-1)d$ . For 18, 11, 4, $-3$ , ...: $a = 18$ and $d = 11 - 18 = -7$ . Then

$t_{33} = 18 + 32(-7) = -206$

The function form gives the same answer: $f(33) = -7(33) + 25 = -206$ (pp. 428–429).

Finding a term when two terms are known. Terms that are $k$ positions apart differ by $k$ common differences. So divide the difference in the terms by the difference in their positions to get $d$ . Suppose $t_7 = 53$ and $t_{11} = 97$ .

The terms differ by $97 - 53 = 44$ , and the positions differ by 4, so $4d = 44$ and $d = 11$ .
Substitute into $53 = a + 6(11)$ to get $a = -13$ .
Then $t_{100} = -13 + 99(11) = 1076$ .

A shortcut is to start at $t_{11}$ and add $d$ another 89 times: $97 + 89(11) = 1076$ . You can also treat the two general-term equations as a linear system and subtract them to find $d$ (pp. 432–433).

Finding the number of terms. Set $t_n$ equal to the last term and solve for $n$ . For 7, 9, 11, ..., 63:

$63 = 7 + (n-1)(2) \;\Rightarrow\; n - 1 = 28 \;\Rightarrow\; n = 29$

The answer must be a whole number. If it is not, the given last number is not a term of the sequence (p. 431).

Modelling with Arithmetic Sequences

Arithmetic sequences describe situations that go up or down by the same amount each period. Simple interest is one example, because the interest is always figured on the original amount. Terry invests $300 in a GIC paying 6% simple interest. He earns$ 300 × 0.06 = $18 each year, so the yearly values are 300, 318, 336, ... with$ a = 300 $and$ d = 18$.

To find when the investment reaches $732, solve$ 732 = 300 + (n-1)(18) $. Then$ 432 = 18(n-1) $, so$ n - 1 = 24 $and$ n = 25 $. The value is$ 732 at the start of the 25th year. A spreadsheet that adds $18 to the previous row gives the same result (pp. 430–431).

Geometric sequences: definition and rules

A geometric sequence is a list of numbers where you multiply by the same number each time to get the next term. That number is the common ratio, $r$ . To find it, divide any term by the term before it (p. 436, p. 439).

In 2, 6, 18, 54, ... each ratio is $6/2 = 18/6 = 54/18 = 3$ , so $r = 3$ .
In 144, 72, 36, 18, ... each ratio is $1/2$ , so $r = 1/2$ .

A sequence is not geometric if the ratios differ. For example, 3, 9, 81, 6561 has ratios 3, 9, 81.

Recursive formula. Start with the first term $a$ . Then multiply by $r$ to get each new term:

$t_1 = a,\quad t_n = r\,t_{n-1}\quad (n>1,\ n\in N)$

For 2, 6, 18, ... this gives $t_1 = 2$ , $t_n = 3t_{n-1}$ (p. 437, p. 439).

General term. Write the terms using $a$ and $r$ : $a,\ ar,\ ar^2,\ ar^3,\ \dots$ The power of $r$ is always one less than the position number, because the first term has not been multiplied by $r$ yet. So

$t_n = ar^{n-1}$

Example: find the 13th term when $a = 9$ and $r = 2$ . You multiply by 2 twelve times, so $t_{13} = 9 \times 2^{12} = 36\,864$ . A table that doubles each term gives the same answer, but it takes much longer (p. 437).

Graphs and the behaviour of $r$

Since $t_n = ar^{n-1}$ is an exponential expression, a geometric sequence is a discrete exponential function, $f(n) = ar^{n-1}$ , with $n \in N$ . Its graph is a set of separate points on an exponential curve, not on a line. Constant ratios (not constant differences) make the curve bend. For example, $f(n) = 16(5)^{n-1}$ gives 16, 80, 400, 2000, ..., which grows very fast (p. 439, p. 445).

The value of $r$ controls what the sequence does:

If $r > 1$ and the terms are positive, the terms increase.
If $0 < r < 1$ , the terms shrink toward zero. For example, multiplying by 0.95 each time makes terms 5% smaller (pp. 438–439).
If $r < 0$ , the sign flips with every multiplication, so the terms alternate between positive and negative. The sequence 5, $-20$ , 80, $-320$ , ... has $r = -4$ (p. 439).

Using geometric sequences to solve problems

Growth and decay by a constant percent. Suppose 3 kg of radioactive material loses 5% each year. Then 95% remains each year, so $a = 3$ and $r = 0.95$ . After 100 years you need the 100th term:

$f(100) = 3(0.95)^{99} \approx 0.019\text{ kg}$

That is about 19 g (p. 438).

Finding the number of terms. Take the sequence 52 612 659, 17 537 553, ..., 11. The ratio is $17\,537\,553 / 52\,612\,659 = 1/3$ . Put the last term into the general term:

$11 = 52\,612\,659\left(\tfrac{1}{3}\right)^{n-1}$

This is hard to solve by hand at this level, so use one of two methods. You can try values of $n$ by guess and check. Or you can graph $y = 52\,612\,659(1/3)^{x-1}$ and $y = 11$ on a graphing calculator and find where they cross. They cross at $x = 15$ , so there are 15 terms (pp. 438–439).

Finding a term when two terms are known. Suppose $t_5 = 45$ and $t_8 = 360$ . There are 3 steps from $t_5$ to $t_8$ , so you multiplied by $r$ three times:

$r^3 = \frac{360}{45} = 8 \Rightarrow r = 2$

To reach $t_{20}$ you need 15 more steps from $t_5$ , so $t_{20} = 45(2)^{15} = 1\,474\,560$ . You never needed the general term (p. 441).

Arithmetic and geometric sequences compared

Both kinds of sequence are built step by step, both have a general term, and both are discrete functions on the natural numbers. The difference is the operation (pp. 442, 445):

An arithmetic sequence adds a fixed amount. Its 1st differences are constant and its graph is points on a line.
A geometric sequence multiplies by a fixed amount. Its ratios are constant and its graph is points on an exponential curve.

Sequences that are neither arithmetic nor geometric

A common misconception is that a sequence with no constant difference and no constant ratio has no rule. That is not true. Many sequences follow other patterns, and differences and ratios often reveal them (p. 435, p. 445).

Example: 1, 8, 16, 26, 39, ... has 1st differences 7, 8, 10, 13 (not constant). Its ratios are also not constant. The 2nd differences are 1, 2, 3, which go up by 1 each time. Continuing the pattern, the next 2nd differences are 4, 5, 6, ... and the next 1st differences are 17, 22, 28, ... So the next terms are $39+17 = 56$ , $56+22 = 78$ , and then 106 (p. 445, p. 448).

The Tower of Hanoi

In this puzzle you move discs of different sizes between three pegs. You move one disc at a time, and a disc may never sit on a smaller one. The goal is to move the whole stack to another peg in the fewest moves (p. 443).

Start with a simpler problem and count moves for 1, 2, and 3 discs. You get 1, 3, 7. For 3 discs you must move the top 2 discs to a spare peg, move the largest disc, and then move the 2 discs back on top of it. In general, moving $n$ discs takes the moves for $n-1$ discs, plus 1 move for the biggest disc, plus the moves for $n-1$ discs again. Each term is double the one before, plus 1:

$t_1 = 1,\quad t_n = 2t_{n-1} + 1$

Applying this repeatedly gives 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023. So 10 discs need at least 1023 moves (p. 444).

A general term. The recursive rule is slow for large $n$ . For 1000 discs you would have to compute 999 terms first. Add 1 to every term:

$2,\ 4,\ 8,\ 16,\ \dots$

These are the powers of 2, a geometric sequence with general term $2^n$ . Subtracting the 1 again gives

$t_n = 2^n - 1$

For $n = 10$ , $2^{10} - 1 = 1023$ , which matches the recursive result (p. 444).

Extending a sequence that is neither arithmetic nor geometric

Some sequences do not add a fixed amount or multiply by a fixed amount. They still may follow a pattern. A good plan is to test the simple types first and then dig deeper.

Step 1: Classify the sequence. Find the 1st differences (each term minus the one before). If they are all equal, the sequence is arithmetic. If not, find the ratios of consecutive terms. If those are all equal, the sequence is geometric. If neither check works, look for another pattern. (pp. 445–446)

Example. Take 1, 8, 16, 26, 39, 56, 78, ... The 1st differences are 7, 8, 10, ..., which are not constant. The ratios are 8 and 2, ..., which are not constant either. So the sequence is neither arithmetic nor geometric. (pp. 445–446)

Step 2: Use higher differences. Take the differences of the differences. For the sequence above, the 1st differences are 7, 8, 10, 13, 17, 22. The 2nd differences are 1, 2, 3, 4, 5. These go up by 1 each time, so a pattern appears. (p. 445)

Step 3: Test the pattern, then work backward. Before trusting a pattern, check that it predicts terms you already know. Then continue it:

The next 2nd differences are 6, 7, 8.
Add each to the last 1st difference (22) to get new 1st differences: 22 + 6 = 28, 28 + 7 = 35, 35 + 8 = 43.
Add each to the last term (78) to get new terms: 78 + 28 = 106, 106 + 35 = 141, 141 + 43 = 184.

The next three terms are 106, 141, and 184. (pp. 445–446)

Finding a recursive rule when ratios are nearly constant

Sometimes the ratios of consecutive terms are not equal but get close to one number, $k$ . This suggests each term is about $k$ times the one before. The rule then needs a small correction. (pp. 446–447)

To find the correction, make a table. Compare each term $t_n$ with $k\,t_{n-1}$ and look at the gap.

Example. For 5, 14, 41, 122, 365, 1094, 3281, ... the 1st differences (9, 27, ...) are not constant, so the sequence is not arithmetic. The ratios are 14/5 = 2.8, 41/14 ≈ 2.93, 122/41 ≈ 2.98, 365/122 ≈ 2.99. They are not equal, so it is not geometric. They are approaching 3, so compare each term with 3 times the term before it:

$3 \times 5 = 15$ , and the next term is 14 (one less)
$3 \times 14 = 42$ , and the next term is 41 (one less)
$3 \times 41 = 123$ , and the next term is 122 (one less)

The pattern holds, so each term is 3 times the previous term minus 1:

$t_1 = 5,\quad t_n = 3t_{n-1} - 1,\quad n > 1,\ n \in N$

(pp. 446–447)

Finding a general term for a sequence of fractions

When the terms are fractions, study the numerators and denominators as two separate sequences. Find a formula for each, then put them together as $t_n = \dfrac{N_n}{D_n}$ . (pp. 447–448)

Example. Take $\frac{3}{4}, \frac{5}{9}, \frac{7}{16}, \frac{9}{25}, \frac{11}{36}, \dots$

Numerators: 3, 5, 7, 9, ... have a common difference of 2, so they are arithmetic. With $a = 3$ and $d = 2$ : $N_n = 3 + (n-1)(2) = 2n + 1$ . (p. 447)
Denominators: 4, 9, 16, 25, ... are the perfect squares $2^2, 3^2, 4^2, 5^2, \dots$ . The first term uses 2, not 1. So the base is one more than the position: $D_n = (n+1)^2$ . (p. 448)

Combining them gives

$t_n = \frac{2n+1}{(n+1)^2},\quad n \in N$

(p. 448)

Do all sequences have a general term?

No. A sequence is an ordered list of numbers, and it does not need a predictable pattern. A sequence has a general term only if some algebraic rule in terms of the position $n$ produces each term. (p. 448)

The prime numbers 2, 3, 5, 7, 11, ... are a clear example. They are well understood, yet no formula or recursive rule is known that generates them. Their spacing is unpredictable. So a sequence can be perfectly well defined without a rule in $n$ . (p. 448)

There is one guarantee. For an arithmetic sequence, $t_n = a + (n-1)d$ always works. For a geometric sequence, $t_n = ar^{n-1}$ always works. For any other kind of sequence, a general term may exist, may be only a recursive rule, or may not exist at all. (p. 448)

Sequences where a term depends on two earlier terms

In the sequences so far, each term depended on at most one earlier term. In a more general recursive sequence, a term can depend on several previous terms. (pp. 451, 453)

The Fibonacci sequence. It began with a rabbit-breeding problem posed by Leonardo Pisano, nicknamed Fibonacci. The number of pairs of rabbits each month gives 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ... Each term is the sum of the two terms before it:

$t_1 = 1,\quad t_2 = 1,\quad t_n = t_{n-1} + t_{n-2},\quad n > 2,\ n \in N$

For example, $t_7 = t_6 + t_5 = 8 + 5 = 13$ . Two starting values are needed because the rule looks back two terms. (pp. 451, 453)

The Lucas sequence. It uses the same rule but starts with $t_1 = 1$ and $t_2 = 3$ . The terms are 1, 3, 4, 7, 11, 18, 29, 47, ... Changing only the starting values gives a different sequence. (pp. 451, 453)

Fibonacci and Lucas sequences behave like a geometric sequence

Take ratios of consecutive Fibonacci terms: 1/1 = 1, 2/1 = 2, 3/2 = 1.5, 5/3 ≈ 1.67, 8/5 = 1.6, and so on. These ratios bounce around but settle close to one value, about 1.618. The Lucas ratios do the same. (pp. 442, 453)

The ratio is never exactly constant, so neither sequence is truly geometric. Both behave much like one, though, with a common ratio near 1.618. (pp. 442, 453)

Why that number? Suppose a geometric sequence $t_n = ar^{n-1}$ followed the Fibonacci rule $t_n = t_{n-1} + t_{n-2}$ . Substituting gives $ar^{n-1} = ar^{n-2} + ar^{n-3}$ . Dividing by $ar^{n-3}$ gives $r^2 = r + 1$ . Solving this equation produces a value of $r$ that you can compare with the ratio found above. (p. 453)

Same relationship, similar properties

Sequences that share the same relationship between consecutive terms have similar properties, even if their starting values differ. Fibonacci and Lucas show this. To test it, pick any two starting numbers, add the previous two terms repeatedly to build a sequence, and compute the ratios of consecutive terms. They should also approach about 1.618. (pp. 442, 453)

Similar relationships also link the two sequences. Adding Fibonacci terms that are two apart gives Lucas numbers. Multiplying a Fibonacci number by the Lucas number in the same position gives another Fibonacci number. (p. 442)

Fibonacci numbers in nature

Fibonacci numbers show up in living things. Many flowers have 3, 5, 8, or 13 petals. Pineapples have spirals, and sunflower heads have spirals of seeds, whose counts are Fibonacci numbers. This is a major reason the sequence is studied beyond the original rabbit problem. (p. 453)

The golden ratio

The golden ratio, written $\varphi$ (Greek letter phi), was known to the ancient Greeks. Euclid defined it using a point C on segment AB so that

$\varphi = \frac{AC}{CB} = \frac{AB}{AC}$

Its value is about 1.618, the same number the Fibonacci ratios approach. The Greeks thought this proportion was especially pleasing to the eye and used it in architecture. It also appears in some body proportions and in the seed spirals of flower heads. (p. 454)

You can see the link by drawing squares with side lengths 1, 1, 2, 3, 5, ... in a growing spiral arrangement. Each new square is placed around the picture, alternating left, up, right, and down. The side lengths follow the Fibonacci sequence. (p. 454)

The Golden Ratio and the Fibonacci Rule

The golden ratio, written $\varphi$ (phi), comes from cutting a segment AB at a point C so that the whole is to the longer part as the longer part is to the shorter part:

$\varphi = \frac{AC}{CB} = \frac{AB}{AC}$

Its value is about 1.618. The ancient Greeks found rectangles with this proportion pleasing and used it in buildings. (p. 454)

The ratios of consecutive Fibonacci terms settle near this same number: $\frac{1}{1}=1$ , $\frac{2}{1}=2$ , $\frac{3}{2}=1.5$ , $\frac{5}{3}\approx1.667$ , $\frac{8}{5}=1.6$ , and so on. The Lucas ratios (3, 1.33, 1.75, 1.57, 1.64, ...) also close in on 1.618. (pp. 451, 454)

Why 1.618? Test a geometric sequence in the Fibonacci rule. A geometric sequence has $t_n = ar^{n-1}$ . Put this into $t_n = t_{n-1} + t_{n-2}$ :

$ar^{n-1} = ar^{n-2} + ar^{n-3}$

Divide every term by $ar^{n-3}$ . This leaves

$r^2 = r + 1, \quad\text{so}\quad r^2 - r - 1 = 0 \quad\Rightarrow\quad r = \frac{1 \pm \sqrt{5}}{2}$

The positive root is about 1.618. So the Fibonacci and Lucas sequences behave roughly like a geometric sequence with $r \approx 1.618$ . They are not exactly geometric, because the starting terms do not fit that pattern perfectly. (p. 453)

The golden ratio gives the same equation. Let $CB = 1$ and $AC = \varphi$ , so $AB = \varphi + 1$ . Then $\frac{AC}{CB} = \frac{AB}{AC}$ becomes $\varphi = \frac{\varphi + 1}{\varphi}$ , which simplifies to $\varphi^2 = \varphi + 1$ . (pp. 453–454)

Linking Fibonacci and Lucas Numbers

The Fibonacci sequence is 1, 1, 2, 3, 5, 8, 13, ... and the Lucas sequence is 1, 3, 4, 7, 11, 18, ... These sequences are closely connected.

Add Fibonacci terms that are two apart. $1 + 2 = 3$ , $1 + 3 = 4$ , $2 + 5 = 7$ . These sums are Lucas numbers.
Multiply a Fibonacci and a Lucas number from the same position. $1 \times 3 = 3$ and $2 \times 4 = 8$ . The products 1, 3, 8, 21, 55, ... are Fibonacci numbers, namely every second one.

(pp. 451–452)

The two sequences have different starting values but the same rule linking each term to the two before it. Sequences built with the same rule tend to share properties. For example, if you pick any two starting numbers and add the previous two terms each time, the ratios still settle near 1.618. You can check this conjecture by generating a new sequence and calculating its ratios. (pp. 452–453)

Series and Partial Sums

A sequence is a list of numbers. A series is the sum of the terms of a sequence. An arithmetic series is the sum of the terms of an arithmetic sequence. The sequence 2, 10, 18, 26, ... has the related series $2 + 10 + 18 + 26 + \dots$ (pp. 458, 461)

A partial sum, $S_n$ , adds only the first $n$ terms:

$S_n = t_1 + t_2 + \dots + t_n$

For 2, 10, 18, 26, ... we get $S_4 = 2 + 10 + 18 + 26 = 56$ . It is "partial" because it covers only part of the series. (pp. 459, 461)

Gauss's Method and the Sum Formulas

To add $1 + 2 + \dots + 100$ , write the sum forward, then write it backward underneath. Add each column. Every column totals 101, and there are 100 columns, so the two rows together equal $100 \times 101 = 10\,100$ . Each row alone is half of that, so the sum is 5050. The method works because the terms go up by a constant amount in one row and down by the same amount in the other, so every pair has the same total. (pp. 458–459)

Applying this to any arithmetic series gives $n$ pairs, each with the sum $t_1 + t_n$ . Halving the total gives:

$S_n = \frac{n(t_1 + t_n)}{2}$

Use this form when you know the first and last terms. For $2 + 10 + 18 + 26$ : $S_4 = \frac{4(2 + 26)}{2} = 56$ . (pp. 459, 461)

Since $t_n = a + (n-1)d$ , you can substitute for $t_n$ and get a second form:

$S_n = \frac{n[2a + (n-1)d]}{2}$

Use this form when you know the first term $a$ and the common difference $d$ . (pp. 459–460)

Example (amphitheatre). There are 50 rows. The first has 23 seats and each row adds 4. The row sizes form an arithmetic sequence with $a = 23$ , $d = 4$ , $n = 50$ :

$S_{50} = \frac{50[2(23) + 49(4)]}{2} = \frac{50(242)}{2} = 6050 \text{ seats}$

(p. 460)

When the Number of Terms Is Unknown

Both formulas need $n$ . If it is not given, find it first by setting the last term equal to the general term, then solve.

Example. Find $-31 - 35 - 39 - \dots - 403$ . Here $a = -31$ and $d = -4$ .

$-403 = -31 + (n-1)(-4)$ $-372 = (n-1)(-4) \quad\Rightarrow\quad n - 1 = 93 \quad\Rightarrow\quad n = 94$

Now the first and last terms are known, so

$S_{94} = \frac{94(-31 + (-403))}{2} = -20\,398$

(pp. 460–461)

Why the Sum Formula Has an $n^2$ Term

The difference between two neighbouring partial sums is one term: $S_n - S_{n-1} = t_n$ . So the 1st differences of the partial sums are the terms of the series. Those terms are arithmetic, so their differences are the constant $d$ . That means the partial sums have constant 2nd differences equal to $d$ .

For 2, 10, 18, 26: the partial sums are 2, 12, 30, 56. The 1st differences are 10, 18, 26, and the 2nd differences are 8, 8, which equals $d$ . Constant 2nd differences mean a quadratic pattern in $n$ , which is why the sum formula expands to include $n^2$ . (p. 459)

Application: Counting Handshakes

Suppose 23 people at a party each shake hands with every other person exactly once. Imagine them arriving one at a time. Each person shakes hands with everyone already there. The handshake counts are 0, 1, 2, ..., 22, which is an arithmetic sequence with $d = 1$ . The total is the series $1 + 2 + \dots + 22$ :

$S_{22} = \frac{22(1 + 22)}{2} = \frac{22 \times 23}{2} = 253$

For $n$ people the same reasoning gives $1 + 2 + \dots + (n-1)$ . (pp. 458–459)

Geometric series

A geometric series is what you get when you add up the terms of a geometric sequence. The sequence 3, 6, 12, 24, ... gives the series 3 + 6 + 12 + 24 + ... (pp. 464, 469).

An ancestor tree is a good example. You have 2 parents, 4 grandparents, 8 great-grandparents, and so on. The counts 2, 4, 8, ... form a geometric sequence with $a = 2$ and $r = 2$ . Adding the generations together gives a geometric series (pp. 464–465).

Deriving the sum formula

Adding many terms by hand is slow, so we find a shortcut. Write the sum of the first $n$ terms:

$S_n = a + ar + ar^2 + \dots + ar^{n-1}$

Now multiply every term by $r$ :

$rS_n = ar + ar^2 + \dots + ar^{n-1} + ar^n$

Write $rS_n$ above $S_n$ so that equal terms line up. Subtract $S_n$ from $rS_n$ . Almost every term cancels. Only $ar^n$ and $-a$ are left:

$(r-1)S_n = ar^n - a$

Divide by $r - 1$ to get the formula (pp. 464–465):

$S_n = \frac{a(r^n - 1)}{r - 1}, \quad r \neq 1$

Lining up equal terms is what makes the cancelling easy to see. This also shows why the sum is tied to an exponential function, since $r^n$ appears in it.

Two forms of the formula

The term $ar^n$ is the $(n+1)$ th term, $t_{n+1}$ . So the formula can also be written (pp. 465–467):

$S_n = \frac{t_{n+1} - t_1}{r - 1}, \quad r \neq 1$

Which form should you use? Start by finding $r$ from two consecutive terms (pp. 467–469).

If you know or can find $n$ , use $\frac{a(r^n-1)}{r-1}$ .
If you know the term after the last one, use $\frac{t_{n+1}-t_1}{r-1}$ .
If the last term is given but $n$ is not, solve $t_n = ar^{n-1}$ for $n$ . A table, spreadsheet, or graphing calculator works well for this.

Example 1 (n is known). A hatchery sees 2, 10, 50, 250 fish hatch on the first four days. The ratios $10/2$ , $50/10$ and $250/50$ all equal 5, so the sequence is geometric. For 10 days, $a = 2$ , $r = 5$ , $n = 10$ :

$S_{10} = \frac{2(5^{10} - 1)}{5 - 1} = 4\,882\,812 \text{ fish}$

(pp. 465–466)

Example 2 (n is unknown). Find $7\,971\,615 + 5\,314\,410 + 3\,542\,940 + \dots + 92\,160$ . The ratio is $r = \frac{5\,314\,410}{7\,971\,615} = \frac{2}{3}$ . A table of $t_n = 7\,971\,615\left(\frac{2}{3}\right)^{n-1}$ shows that 92 160 is term 12. The next term is 61 440. Then:

$S = \frac{61\,440 - 7\,971\,615}{\frac{2}{3} - 1} = 23\,730\,525$

Using $a$ , $r$ and $n = 12$ in the other formula gives the same answer (pp. 467–468).

Ancestor tree. For 13 generations, $a = 2$ , $r = 2$ , $n = 13$ :

$S_{13} = \frac{2(2^{13}-1)}{2-1} = 16\,382 \text{ people}$

(pp. 464–465)

The special case r = 1

Both formulas have $r - 1$ in the denominator. If $r = 1$ that means dividing by zero, so the formulas fail. But this case is easy anyway. Every term equals $a$ , so the series is $a + a + a + \dots$ with $n$ terms. The sum is $S_n = na$ (p. 470).

Pascal's triangle

Pascal's triangle is a triangle of numbers. Each row starts and ends with 1. Every other entry is the sum of the two entries just above it (pp. 462, 465):

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

The marble-toy paths follow the same rule. The number of paths to a bin is the sum of the paths to the two bins above it.

Using the triangle to expand binomials

The top row, a single 1, is row 0. The numbers in row $n$ are the coefficients of $(a+b)^n$ (pp. 463, 465). For example, $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$ uses row 3: 1, 3, 3, 1.

The exponents follow a pattern too (pp. 463, 465):

The exponents of the first term start at $n$ and fall to 0.
The exponents of the second term start at 0 and rise to $n$ .
In every term, the two exponents add to $n$ .
There are $n + 1$ terms.

For $(x+y)^6$ the terms are $x^6,\ 6x^5y,\ 15x^4y^2,\ 20x^3y^3,\ 15x^2y^4,\ 6xy^5,\ y^6$ .

Example 3: $(x-2)^5$ . Row 5 is 1, 5, 10, 10, 5, 1. Use $x$ and $-2$ as the two terms:

$x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32$

Powers of $-2$ alternate in sign, so the signs of the terms alternate (p. 464).

Example 4: $(5x+2y)^3$ . Treat $5x$ and $2y$ as whole units. Row 3 is 1, 3, 3, 1:

$(5x)^3 + 3(5x)^2(2y) + 3(5x)(2y)^2 + (2y)^3 = 125x^3 + 150x^2y + 60xy^2 + 8y^3$

A common mistake is to forget to raise the 5 and the 2 to the powers (pp. 464–465).

Patterns inside Pascal's triangle

The outer diagonals are all 1s.
The next diagonal is the counting numbers 1, 2, 3, ..., an arithmetic sequence with $d = 1$ .
The next diagonal is the triangular numbers 1, 3, 6, 10, ..., where $t_1 = 1$ and $t_n = t_{n-1} + n$ . For example, $6 + 4 = 10$ . These match the triangle stacks of golf balls.

(p. 465)

Check yourself

An arithmetic sequence has first term 7 and common difference 4. Find $t_{30}$ and the sum of its first 30 terms.
Find the sum of the first 8 terms of $6 + 18 + 54 + \dots$ . Which formula did you choose, and why?
How can you tell whether a sequence is arithmetic, geometric, or neither? What would you check next if it is neither?
The Fibonacci rule is $t_n = t_{n-1} + t_{n-2}$ . What number do the ratios of consecutive terms settle near?
Expand $(2x - 1)^4$ using Pascal's triangle.

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Functions 11

Chapter 7: Discrete Functions: Sequences and Series

Sequences, Terms and Recursive Rules

Sequences as Discrete Functions

Arithmetic Sequences

Finding Terms of an Arithmetic Sequence

Modelling with Arithmetic Sequences

Geometric sequences: definition and rules

Graphs and the behaviour of rrr

Using geometric sequences to solve problems

Arithmetic and geometric sequences compared

Sequences that are neither arithmetic nor geometric

The Tower of Hanoi

Extending a sequence that is neither arithmetic nor geometric

Finding a recursive rule when ratios are nearly constant

Finding a general term for a sequence of fractions

Do all sequences have a general term?

Sequences where a term depends on two earlier terms

Fibonacci and Lucas sequences behave like a geometric sequence

Same relationship, similar properties

Fibonacci numbers in nature

The golden ratio

The Golden Ratio and the Fibonacci Rule

Linking Fibonacci and Lucas Numbers

Series and Partial Sums

Gauss's Method and the Sum Formulas

When the Number of Terms Is Unknown

Why the Sum Formula Has an n2n^2n2 Term

Application: Counting Handshakes

Geometric series

Deriving the sum formula

Two forms of the formula

The special case r = 1

Pascal's triangle

Using the triangle to expand binomials

Patterns inside Pascal's triangle

Check yourself

Chapter 7: Discrete Functions: Sequences and Series

Sequences, Terms and Recursive Rules

Sequences as Discrete Functions

Arithmetic Sequences

Finding Terms of an Arithmetic Sequence

Modelling with Arithmetic Sequences

Geometric sequences: definition and rules

Graphs and the behaviour of rrr

Using geometric sequences to solve problems

Arithmetic and geometric sequences compared

Sequences that are neither arithmetic nor geometric

The Tower of Hanoi

Extending a sequence that is neither arithmetic nor geometric

Finding a recursive rule when ratios are nearly constant

Finding a general term for a sequence of fractions

Do all sequences have a general term?

Sequences where a term depends on two earlier terms

Fibonacci and Lucas sequences behave like a geometric sequence

Same relationship, similar properties

Fibonacci numbers in nature

The golden ratio

The Golden Ratio and the Fibonacci Rule

Linking Fibonacci and Lucas Numbers

Series and Partial Sums

Gauss's Method and the Sum Formulas

When the Number of Terms Is Unknown

Why the Sum Formula Has an n2n^2n2 Term

Application: Counting Handshakes

Geometric series

Deriving the sum formula

Two forms of the formula

The special case r = 1

Pascal's triangle

Using the triangle to expand binomials

Patterns inside Pascal's triangle

Check yourself

Graphs and the behaviour of $r$

Why the Sum Formula Has an $n^2$ Term

Graphs and the behaviour of $r$

Why the Sum Formula Has an $n^2$ Term