3 SequencesIn mathematics, a sequence is a succession of numbers, called terms, that follow a given rule. For example:9, 16, 25, 36, 49, …is a sequence of square numbers starting with 9.A sequence can be infinite, as shown by the … at the end of the sequence shown above, or it can be finite. For example:3, 6, 12, 24, 48, 96is a finite sequence containing six terms.A sequence can be defined by:a formula for the nth term of the sequence, ora recurrence relation together with the first term of the sequence.
5 The formula for the nth term The nth term, or the general term, of a sequence is often given using superscript (or suffix) notation as un.The 1st term is then called u1,the 2nd term is u2,the 3rd term is u3,the 4th term is u4,the 5th term is u5 and so on.Letters other than u can be used. For example, the terms in a sequence could also be given by t1, t2, t3, t4, … tn.Any term in a sequence can be found by substituting its position number into a given formula for un.
6 The formula for the nth term For example, the formula for the nth term of a sequence is given by un = 4n – 5.Find the first five terms in the sequence.u1 =4 × 1 – 5 =–1u2 =4 × 2 – 5 =3u3 =4 × 3 – 5 =7u4 =4 × 4 – 5 =11This can be compared to using function notation, where instead of using f(x) we use un, where x and n are variables. The important difference is that, in general, the variable x can be any real number, whereas n can only be a positive whole number.u5 =4 × 5 – 5 =15The first five terms in the sequence are: –1, 3, 7, 11 and 15.
9 Recurrence relationsThis sequence can also be defined by a recurrence relation.To define a sequence using a recurrence relation we need the value of the first term and an expression relating each term to a previous term.For the sequence –1, 3, 7, 11, 15, …, each term can be found by adding 4 to the previous term.We can write:u1 = –1u2 = u1 + 4 = 3u3 = u2 + 4 = 7The recurrence relation given here can be interpreted as ‘each term is equal to the previous term plus 4’.u4 = u3 + 4 = 11 and so on.In general:un+1 = un + 4
10 Recurrence relationsA recurrence relation together with the first term of a sequence is called an inductive definition.So the inductive definition for the sequence –1, 3, 7, 11, 15, … is u1 = –1, un+1 = un + 4.A sequence is given by the recurrence relation un+1 = 2un + 1 with u1 = 3. Write down the first five terms of the sequence.u1 = 3u2 = (2 × 3) + 1 =7u3 = (2 × 7) + 1 =15Explain that a recurrence relation alone cannot be used to define a sequence; the first term or terms must also be given. For example, the sequence 4, 8, 12, 16, … and the sequence 1, 5, 9, 13, … are also defined by the recurrence relation un+1 = un + 4. The difference is that they have different values for the first term.u4 = (2 × 15) + 1 =31u5 = (2 × 31) + 1 =63So the first five terms in the sequence are 3, 7, 15, 31 and 63.
11 Using an inductive definition Use this activity to demonstrate how an inductive definition can be used to generate a sequence.
13 Arithmetic sequencesIn an arithmetic sequence (or arithmetic progression) the difference between any two consecutive terms is always the same. This is called the common difference.For example, the sequence:8, 11, 14, 17, 20, …is an arithmetic sequence with 3 as the common difference.We could write this sequence as:8,8 + 3,,,, …You may wish to add that if the terms of an arithmetic sequence were plotted on a graph they would lie on a straight line.or8,8 + 3,8 + (2 × 3),8 (3 × 3),8 + (4 × 3), …
14 Arithmetic sequencesIf we call the first term of an arithmetic sequence a and the common difference d we can write a general arithmetic sequence as:a,a + d,a + 2d,a + 3d,a + 4d, …The nth term of an arithmetic sequence with first term a and common difference d isa + (n – 1)dAlso:The formula for the nth term is sometimes called a deductive definition, since it defines the value of un directly, while an inductive definition defines each term in relation to a previous term.The inductive definition of an arithmetic sequence with first term a and common difference d isu1 = a, un+1 = un + d
15 Arithmetic sequencesWhat is the formula for the nth term of the sequence 10, 7, 4, 1, –2 …?This is an arithmetic sequence with first term a = 10 and common difference d = –3.The nth term is given by a + (n – 1)d so:un = 10 – 3(n – 1)= 10 – 3n + 3= 13 – 3nLet’s check this formula for the first few terms in the sequence:u1 = 13 – 3 × 1 = 10u2 = 13 – 3 × 2 = 7u3 = 13 – 3 × 3 = 4
16 Arithmetic sequencesFind the number of terms in the finite arithmetic sequence –7, –1, 5, … 71.This is an arithmetic sequence with first term a = –7 and common difference d = 6.The nth term is given by a + (n – 1)d so:un = –7 + 6(n – 1)= –7 + 6n – 6= 6n – 13We can find the value of n for the last term by solving:6n – 13 = 716n = 84n = 14So, there are 14 terms in the sequence.
17 Arithmetic sequencesThe 4th term in an arithmetic sequence is 12 and the 20th term is 92. What is the formula for the nth term of this sequence?Using the 4th term:a + 3d = 12Using the 20th term:a + 19d = 92Subtracting the first equation from the second equation gives:16d = 80d = 5Substitute this into the first equation:a + 15 = 12Explain that we can find the values of the first term a and common difference d by using the information given to write a pair of simultaneous equations.a = –3The nth term of an arithmetic sequence with a = –3 and d = 5 is:un = –3 + 5(n –1)= –3 + 5n – 5= 5n – 8
19 Series The sum of all the terms of a sequence is called a series. For example:1, 3, 5, 7, 9, … is a sequencewhile:… is a series.When the difference between each term in a series is constant, as in this example, the series is called an arithmetic series or arithmetic progression (AP for short).The sum of a series containing n terms is often denoted by Sn, so for the series given above we could write:S5 =Point out that, as for a sequence, a series can be finite, coming to an end after a given number of terms, or it can be infinite, continuing indefinitely.= 25When n is large, a more systematic approach for calculating the sum of a given number of terms is required.
20 Gauss’ methodIt is said that when the famous mathematician Karl Friedrich Gauss was a young boy at school, his teacher asked the class to add together every whole number from one to a hundred.The teacher expected this activity to keep the class occupied for some time and so he was amazed when Gauss put up his hand and gave the answer, 5050, almost immediately!
21 Gauss’ methodGauss worked the answer out by noticing that you can quickly add together consecutive numbers by writing the numbers once in order and once in reverse order and adding them together.So to add the numbers from 1 to 100:1+2345…9899100S =100+99989796…321S =2S =101+101+101+101+101+…101+101+101So:2S = 100 × 101=S = 5050
23 The sum of the first n natural numbers To find the sum of the first n natural numbers we can generalize Gauss’ method as follows.Write the sum of the first n natural numbers as:1+23…(n – 2)(n –1)nS =n+(n –1)(n – 2)…321S =2S =(n + 1)+(n + 1)+(n + 1)+…(n + 1)+(n + 1)+(n + 1)This gives us:2S = n(n + 1)So:The sum of the first n natural numbers is given by
24 The sum of the first n natural numbers What is the sum of the first 30 natural numbers?… + 30 == 465What is the sum of the natural numbers from 21 to 30?… + 30 =( … + 30) – ( … + 20)For the second example establish that the sum of the natural numbers from 21 to 30 can be found be subtracting the sum of the natural numbers from 1 to 20 from the sum of the natural numbers from 1 to 30.= 465 – 210= 255
26 The sum of an arithmetic series Gauss’ method can be applied to any arithmetic series of the general forma + (a + d) + (a + 2d) + (a +3d) + … + (a + (n – 1)d)where a is the first term in the series, d is the common difference and n is the number of terms.Let’s call the last term l so that:l = (a + (n – 1)d)The sum of the first n terms can now be written as:a+…(l – 2d)(l – d)lSn =(a + d)(a + 2d)l+…(a + 2d)(a + d)aSn =(l – d)(l – 2d)2Sn=(a + l)+(a + l)+(a + l)+…(a + l)+(a + l)+(a + l)
27 The sum of an arithmetic series This gives us:2Sn = n(a + l)So:The sum of the first n terms in an arithmetic series iswhere a is the first term and l is the last.If the last term is not known this formula can be written in terms of a and n by substituting (a + (n – 1)d) for l in the above.The first form can be remembered as the number of terms multiplied by the mean term.An alternative formula for the sum of an arithmetic series is then:
28 The sum of an arithmetic series Find the sum of the first 20 terms of the arithmetic series…We don’t know the last term so we can use:with a = 5, d = 6 and n = 20.An alternative could be to write the nth term of the series in the form 6n – 1 (using a + (n – 1)d or otherwise). The last term is then 6 × 20 – 1 = 119.The sum can then be found using n/2(a + l).S20 = 10( )= 1240
31 Using Σ notationWhen working with series, the Greek symbol Σ (the capital letter sigma) is used to mean ‘the sum of’.For example:… and this is the last value of r.This is the first value of r …represents a finite series containing n terms:u1 + u2 + u3 + … + unAdd that we use sigma, the Greek letter S, to stand for sum.The terms in the series are obtained by substituting 1, 2, 3, …, n in turn for r in ur.
32 Using Σ notationFor example, suppose we want to find the sum of the first 4 terms of the series whose nth term is of the form 3n – 1.We can write:(3 × 1 – 1) + (3 × 2 – 1) + (3 × 3 – 1) + (3 × 4 – 1)=The initial value of r doesn’t have to be 1. For example:Draw students’ attention to the fact that the first example contains 4 terms when r takes integral values between 1 and 4. When r takes integral values between 3 and 8, as in the second example, there are 6 terms. Conclude that the number of terms in a series can be found by subtracting the lowest value of r from the highest value of r and adding 1.Infinite series are given by writing an ∞ symbol above the Σ.For example:
34 Using Σ notation Evaluate Substituting r = 2, 3, 4, …,15 into 25 – 2r gives us the arithmetic series … + –5.There are 14 terms in this sequence because both r = 2 and r = 15 are included.We can evaluate this by putting a = 21, l = –5 and n = 14 into the formulaExplain that we can find the first and last terms by substituting r =2 and r = 15 into 25 – 2r.To find n, the number of terms in the series, we find the difference between 15 and 2 and add 1. This is because both the 2nd and the 15th term are included.So:= 112
36 Examination-style question The sum of the first 3 terms of an arithmetic series is 21 and the sum of the next three terms is 66.Find the value of the first term and the common difference.Write an expression for the nth term of the series un.Find the sum of the first 10 terms.a) The sum of the first 3 terms can be written as:a + (a + d) + (a + 2d) = 3a + 3d3a + 3d = 21Soa + d = 71The sum of the next 3 terms can be written as:(a + 3d) + (a + 4d) + (a + 5d) = 3a + 12d3a + 12d = 66Soa + 4d = 222
37 Examination-style question 2– :13d = 15d = 5a = 2b) In general, for an arithmetic series un = a + (n – 1)d soun = 2 + 5(n – 1)= 5n – 3c)u10 = (5 ×10) – 3= 47Now using the formula with a = 2 and l = 47:= 245