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Chapter Overview
Determine the value of 2 + 4 + 6 +β―+ 100
2:: Arithmetic Series
The first term of a geometric sequence is 3 and the second term 1. Find the sum to infinity.
3:: Geometric Series
Determine the value of
π=1
100
(3π + 1)
4:: Sigma Notation
If π1 = π and ππ+1 = 2ππ β 1, determine π3 in terms of π.
5:: Recurrence Relations
NEW TO A LEVEL 2017!Identifying whether a sequence is increasing, decreasing, or periodic.
1:: Sequences
Sequences
2, 5, 8, 11, 14, β¦
+3 +3 +3 This is a:
Arithmetic Sequence?
Geometric Sequence(We will explore these later in the chapter)
?3, 6, 12, 24, 48, β¦
Γ 2 Γ 2 Γ 2
common difference π
common ratio π
?
?An arithmetic sequence is one which has a common difference between terms.
A sequence is an ordered set of mathematical objects. Each element in the sequence is called a term.
Sequences
1, 1, 2, 3, 5, 8, β¦This is the Fibonacci Sequence. The terms follow a recurrence relation because each term can be generated using the previous ones.We will encounter recurrence relations later in the chapter.
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The fundamentals of sequences
π’π The πth term. So π’3 would refer to the 3rd term.
π The position of the term in the sequence.
2, 5, 8, 11, 14,β¦
π = 3 π’3 = 8
? Meaning
? Meaning
? ??
πth term of an arithmetic sequence
We use π to denote the first term. π is the difference between terms, and π is the position of the term weβre interested in. Therefore:
1st Term 2nd Term 3rd Term πth term. . .
π π + π π + 2π π + (π β 1)π? ? ? ?
! πth term of arithmetic sequence:
π’π = π + π β 1 π
. . .
An arithmetic sequence is one which has a common difference between terms.
πth term of an arithmetic sequence
! πth term of arithmetic sequence:
π’π = π + π β 1 π
Example 1
The πth term of an arithmetic sequence is π’π = 55 β 2π.a) Write down the first 3 terms of the
sequence.b) Find the first term in the sequence that is
negative.
a) ππ = ππ β π π = ππ, ππ = ππ, ππ = ππb) ππ β ππ < π β π > ππ. π β΄ π = πππππ = ππ β π ππ = βπ
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Example 2
Find the πth term of each arithmetic sequence.a) 6, 20, 34, 48, 62b) 101, 94, 87, 80, 73
a) π = π,π = ππππ = π + ππ π β π= πππ β π
b) π = πππ, π = βπππ = πππ β π π β π= πππ β ππ
Tip: Always write out π =, π =, π = first.
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Further Examples
[Textbook] A sequence is generated by the formula π’π = ππ + π where π and π are constants to be found.Given that π’3 = 5 and π’8 = 20, find the values of the constants π and π.
π’3 = 5 β π + 2π = 5π’8 = 20 β π + 7π = 20
Solving simultaneously, π = 3, π = β4
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Remember that an arithmetic sequence is one where there is a common difference between terms.
ππ + π = πππ β ππ
πππ β πππ + π = π β ππ β π π β π = π
π =π
π, π = π
For which values of π₯ would the expression β8, π₯2 and 17π₯ form the first three terms of an arithmetic sequence.
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Test Your Understanding
Edexcel C1 May 2014(R) Q10
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Exercise 3A
Pearson Pure Mathematics Year 2/ASPages 61-62
Extension[STEP I 2004 Q5] The positive integers can be split into five distinct arithmetic progressions, as shown:
A: 1, 6, 11, 16, β¦B: 2, 7, 12, 17, β¦C: 3, 8, 13, 18, β¦D: 4, 9, 14, 19, β¦E: 5, 10, 15, 20, β¦
Write down an expression for the value of the general term in each of the five progressions. Hence prove that the sum of any term in B and any term in C is a term in E.
Prove also that the square of every term in B is a term in D. State and prove a similar claim about the square of every term in C.
i) Prove that there are no positive integers π₯ and π¦ such that π₯2 + 5π¦ = 243723
ii) Prove also that there are no positive integers π₯and π¦ such π₯4 + 2π¦4 = 26081974
1
A: ππ β π B: ππ β π C: ππ β πD: ππ β π E: ππ
Term in B + Term in C (note that the positions in each sequence could be different, so use π and π):
ππ β π + ππβ π= π(π +π β π)
hence a term in E.
Square of term in B:
ππ β π π = ππππ β πππ + π
= π πππ β ππ + π β π
hence a term in D.
Square of term in C:
ππ β π π = ππππ β πππ + π
= π πππ β ππ + π β π
hence also a term in D.
i) 243723 is a term in C. Note that whatever sequence ππ is in, adding ππwill keep it in the same sequence as it is a multiple of 5. So we need to prove no term squared can give a term in C.We have already proved a term in B or in C, squared, gives a term in D. We can similarly show a term in A, squared, stays in A, and a term in E squared stays in E. Finally, a term in D squared gives a term in A. Since the sequences A, B, C, D, E clearly span all positive integers, there cannot be any integer solutions to the equation.
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Series
A series is a sum of terms in a sequence.You will encounter βseriesβ in many places in A Level:
Arithmetic Series (this chapter!)Sum of terms in an arithmetic sequence.
2 + 5 + 8 + 11
Taylor Series (Further Maths)Expressing a function as an infinite series, consisting of polynomial terms.
ππ₯ = 1 +π₯
1!+π₯2
2!+π₯3
3!+β―
Extra Notes: A βseriesβ usually refers to an infinitesum of terms in a sequence. If we were just summing some finite number of them, we call this a partial sum of the series.
e.g. The βHarmonic Seriesβ is
1 +1
2+
1
3+
1
4+β― , which is
infinitely many terms. But a we could get a partial sum,
e.g. π»3 = 1 +1
2+
1
3=
11
6
However, in this syllabus, the term βseriesβ is used to mean either a finite or infinite addition of terms.
Binomial Series (Later in Year 2)You did Binomial expansions in Year 1. But when the power is negative or fractional, we end up with an infinite series.
1 + π₯ = 1 + π₯12 = 1 +
1
2π₯ β
1
8π₯2 +
1
6π₯3 ββ―
Terminology: A βpower seriesβ is an infinite polynomial with increasing powers of π₯. There is also a chapter on power series in the Further Stats module.
https://www.youtube.com/watch?v=Prc7h8lyDXg
Arithmetic Series
π’π = π + π β 1 π ππ =π
22π + π β 1 π
πth term ! Sum of first π terms
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Letβs prove it!
ππ = π + π + π + π + 2π +β― + π + π β 1 πππ = π + π β 1 π + + π + 2π + π + π + πAdding:2ππ = 2π + π β 1 π +β―+ 2π + π β 1 π = π 2π + π β 1 π
β΄ ππ =π
22π + π β 1 π
Example:
Take an arithmetic sequence 2, 5, 8, 11, 14, 17,β¦π5 = 2 + 5 + 8 + 11 + 14
Reversing:π5 = 14 + 11 + 8 + 5 + 2
Adding these:2π5 = 16 + 16 + 16 + 16 + 16 = 16 Γ 5 = 80β΄ π5 = 40
Proving more generally:
The idea is that each pair of terms, at symmetrically opposite ends, adds to the same number.
Exam Note: The proof has been an exam question before. Itβs also a university interview favourite!
Alternative Formula
π + π + π +β―+ πΏ
Suppose last term was πΏ.We saw earlier that each opposite pair of terms (first and last, second and second last, etc.) added to the same total, in this case π + πΏ.
There are π
2pairs, therefore:
!
ππ =π
2π + πΏ
Examples
Find the sum of the first 30 terms of the following arithmetic sequencesβ¦
1 2 + 5 + 8 + 11 + 14β¦ π30 = 1365
100 + 98 + 96 +β― π30 = 2130
π + 2π + 3π +β― π30 = 465π
2
3
?
?
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Tips: Again, explicitly write out "π =β―, π = β― , π = β―β. Youβre less likely to make incorrect substitutions into the formula.
Make sure you write ππ = β― so you make clear to yourself (and the examiner) that youβre finding the sum of the first π terms, not the πth term.
πΊπ > ππππ, π = π, π = ππ
πππ + π β π π > ππππ
π
ππ + π β π π > ππππ
πππ + ππ β ππππ > ππ < βππ. π ππ π > ππ. π
So 28 terms needed.
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Find the greatest number of terms for the sum of 4 + 9 + 14 +β― to exceed 2000.
Edexcel C1 Jan 2012 Q9
π = 400
π = Β£24450
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Test Your Understanding
Exercise 3BPearson Pure Mathematics Year 2/ASPages 64-66
[MAT 2008 1I]The function π π is defined for positive integers π by
π π = sum of digits of πFor example, π 723 = 7 + 2 + 3 = 12.The sum
π 1 + π 2 + π 3 +β―+ π 99equals what?
Extension
1
3
[MAT 2007 1J]The inequality
π + 1 + π4 + 2 + π9 + 3 + β―+ π10000 + 100 > π
Is true for all π β₯ 1. It follows thatA) π < 1300B) π2 < 101C) π β₯ 10110000
D) π < 5150
π will be largest when π is its smallest, so let π = π.Each of the π terms are therefore 1, giving the LHS:
πππ+ π+ π + π + β―+ πππ
πππ+πππ
ππ + ππ Γ π = ππππ
The answer is D.
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[AEA 2010 Q2]The sum of the first π terms of an arithmetic series is π and the sum of the first π terms of the same arithmetic series is π, where π and π are positive integers and π β π.Giving simplified answers in terms of π and π, finda) The common difference of the terms in this
series,b) The first term of the series,c) The sum of the first π + π terms of the
series.
Solution on next slide.
When we sum all digits, we can separately consider sum of all units digits, and the sum of all tens digits.Each of 1 to 9 occurs ten times as units digit, so sum is ππ Γ π+ π+ β―+ π = πππSimilarly each of 1 to 9 occurs ten times as tens digit, thus total is πππ + πππ = πππ.
2
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Solution to Extension Q2
[AEA 2010 Q2]The sum of the first π terms of an arithmetic series is π and the sum of the first π terms of the same arithmetic series is π, where π and π are positive integers and π β π.Giving simplified answers in terms of π and π, finda) The common different of the terms in this series,b) The first term of the series,c) The sum of the first π + π terms of the series.
Recap of Arithmetic vs Geometric Sequences
2, 5, 8, 11, 14, β¦
+3 +3 +3 This is a:
Arithmetic Sequence?
Geometric Sequence?3, 6, 12, 24, 48, β¦
Γ 2 Γ 2 Γ 2
common difference π
common ratio π
?
?
! A geometric sequence is one in which there is a common ratio between terms.
Quickfire Common Ratio
Identify the common ratio π:
1, 2, 4, 8, 16, 32,β¦ π = 21
27, 18, 12, 8,β¦ π = 2/32
10, 5, 2.5, 1.25,β¦ π = 1/23
5,β5, 5, β5, 5, β5,β¦ π = β14
π₯,β2π₯2, 4π₯3 π = β2π₯5
1, π, π2, π3, β¦ π = π6
4,β1, 0.25,β0.0625,β¦ π = β0.257
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An alternating sequence is one which oscillates between positive and negative.
πth term
Arithmetic Sequence Geometric Sequence
π’π = π + π β 1 π π’π = πππβ1
Determine the 10th and πth terms of the following:
3, 6, 12, 24, β¦
40, -20, 10, -5, β¦
? ?
π = 3, π = 2π’10 = 3 Γ 29 = 1536π’π = 3 2πβ1
π = 40, π = β1
2
π’10 = 40 Γ β1
2
9
= β5
64
π’π = β1 πβ1 Γ5
2πβ4
Tip: As before, write out π =and π = first before substituting.
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Further Example
[Textbook] The second term of a geometric sequence is 4 and the 4th
term is 8. The common ratio is positive. Find the exact values of:a) The common ratio.b) The first term.c) The 10th term.
π’2 = 4 β ππ = 4 (1)π’4 = 8 β ππ3 = 8 (2)
a) Dividing (2) by (1) gives π2 = 2, π π π = 2
b) Substituting, π =4
π=
4
2= 2 2
c) π’10 = ππ9 = 64
Tip: Explicitly writing π’2 = 4 first helps you avoid confusing the πth term with the βsum of the first π termsβ (the latter of which weβll get onto).
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Further Example
[Textbook] The numbers 3, π₯ and π₯ + 6 form the first three terms of a positive geometric sequence. Find:
a) The value of π₯.b) The 10th term in the sequence.
π₯
3=π₯ + 6
π₯β π₯2 = 3π₯ + 18
π₯2 β 3π₯ β 18 = 0π₯ + 3 β6 = 0π₯ = 6 ππ β 3But there are no negative terms so π₯ = 6
π =π₯
3=6
3= 2 π = 3 π = 10
π’10 = 3 Γ 29 = 1536
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Hint: Youβre told itβs a geometric sequence, which means the ratio between successive terms must be the
same. Consequently π’2
π’1=
π’3
π’2
Exam Note: This kind of question has appeared in the exam multiple times.
πth term with inequalities
[Textbook] What is the first term in the geometric progression 3, 6, 12, 24,β¦ to exceed 1 million?
π’π > 1000000 π = 3, π = 2β΄ 3 Γ 2πβ1 > 1000000
2πβ1 >1000000
3
log 2πβ1 > log1000000
3
π β 1 >log
10000003
πππ2π β 1 > 18.35π > 19.35π = 20
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Test Your Understanding
All the terms in a geometric sequence are positive.The third term of the sequence is 20 and the fifth term 80. What is the 20th term?
π’3 = 20 β ππ2 = 20π’5 = 80 β ππ4 = 80
Dividing: π2 = 4, β΄ π = 2
π =20
π2=20
4= 5
π’20 = 5 Γ 219 = 2621440
The second, third and fourth term of a geometric sequence are the following:π₯, π₯ + 6, 5π₯ β 6
a) Determine the possible values of π₯.b) Given the common ratio is positive, find the common ratio.c) Hence determine the possible values for the first term of the sequence.
π₯ + 6
π₯=5π₯ β 6
π₯ + 6π₯ + 6 2 = π₯ 5π₯ β 6π₯2 + 12π₯ + 36 = 5π₯2 β 6π₯4π₯2 β 18π₯ β 36 = 02π₯2 β 9π₯ β 18 = 02π₯ + 3 π₯ β 6
π₯ = β3
2ππ π₯ = 6
π =π₯ + 6
π₯=6 + 6
6= 2
π’2 = π₯ = 6ππ = 6
π =6
2= 3
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Exercise 3C
Pearson Pure Mathematics Year 2/ASPages 69-70
Sum of the first π terms of a geometric series
Arithmetic Series Geometric Series
ππ =π
22π + π β 1 π ππ =
π 1 β ππ
1 β π? ?
Proof: ππ = π + ππ + ππ2 +β―+ πππβ2 + πππβ1
Mutiplying by π:πππ = ππ + ππ2 +β― + πππβ1 + πππ
Subtracting:ππ β πππ = π β πππ
ππ 1 β π = π 1 β ππ
ππ =π 1 β ππ
1 β π
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Exam Note: This once came up in an exam. And again is a university interview favourite!
Examples
ππ =π 1 β ππ
1 β π
3, 6, 12, 24, 48,β¦
Find the sum of the first 10 terms.
π = 3, π = 2, π = 10
ππ =3 1 β 210
1 β 2= 3069
4, 2,1,1
2,1
4,1
8, β¦ π = 4, π =
1
2, π = 10
ππ =
4 1 β12
10
1 β12
=1023
128
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Geometric Series
Harder Example
Find the least value of π such that the sum of 1 + 2 + 4 + 8 +β―to π terms would exceed 2 000 000.
π = 1, π = 2, π =?ππ > 2 000 000
1 1 β 2π
1 β 2> 2 000 000
2π β 1 > 2 000 0002π > 2 000 001π > log2 2000001π > 20.9
So 21 terms needed.
πππ2 both sides to cancel out β2 to the power ofβ.
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Test Your Understanding
Edexcel C2 June 2011 Q6
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Exercise 3D
Pearson Pure Mathematics Year 2/ASPages 72-73
Extension
[MAT 2010 1B]The sum of the first 2π terms of
1, 1, 2,1
2, 4,
1
4, 8,
1
8, 16,
1
16, β¦
is
A) 2π + 1 β 21βπ
B) 2π + 2βπ
C) 22π β 23β2π
D)2πβ2βπ
3
1 There are interleaved sequences. If we want ππterms, we want π terms of π, π, π, π, β¦ and π
terms of π,π
π,π
π, β¦
For first sequence:
πΊπ =π π β ππ
π β π= ππ β π
For second sequence:
πΊπ =π π β
ππ
π
π βππ
=π β πβπ
π/π= π β ππβπ
Therefore the sum of both is (A).
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Divergent vs Convergent
1 + 2 + 4 + 8 + 16 + ...
What can you say about the sum of each series up to infinity?
1 β 2 + 3 β 4 + 5 β 6 + β¦
1 + 0.5 + 0.25 + 0.125 + ...
1
1+1
2+1
3+1
4+β―
This is divergent β the sum of the values tends towards infinity.
This is divergent β the running total alternates either side of 0, but gradually gets further away from 0.
This is convergent β the sum of the values tends towards a fixed value, in this case 2.
This is divergent . This is known as the Harmonic Series
1
1+1
4+1
9+
1
16+β―
This is convergent . This is known as the Basel Problem, and the value is π2/6.
Definitely NOT in the A Level syllabus, and just for fun...
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Sum to Infinity
1 + 0.5 + 0.25 + 0.125 + ...
1 + 2 + 4 + 8 + 16 + ...
Why did this infinite sum converge (to 2)β¦
β¦but this diverge to infinity?
The infinite series will converge provided that β1 < π < 1 (which can be written as π < 1), because the terms will get smaller.
Provided that π < 1, what happens to ππ as π β β?
For example π
π
ππππππis very close to 0.
We can see that as π β β, ππ β π.
How therefore can we use the ππ =π 1βππ
1βπformula
to find the sum to infinity, i.e. πβ?
! For a convergent geometric series,
πβ =π
1 β π
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! A geometric series is convergent if π < 1.
Quickfire Examples
1,1
2,1
4,1
8, β¦ π = π, π =
π
ππΊβ = π
27,β9,3,β1,β¦ π = ππ, π = βπ
ππΊβ =
ππ
π
π, π2, π3, π4, β¦ π = π, π = π πΊβ =π
π β ππ€βπππ β 1 < π < 1
? ?
? ?
?
??
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π, 1,1
π,1
π2, β¦ π = π, π =
π
ππΊβ =
ππ
π β π???
Further Examples
[Textbook] The fourth term of a geometric series is 1.08 and the seventh term is 0.23328.a) Show that this series is convergent.b) Find the sum to infinity of this
series.
π’4 = 1.08 β ππ3 = 1.08π’7 = 0.23328 β ππ6 = 0.23328
Dividing:π3 = 0.216 β΄ π = 0.6
The series converges because 0.6 < 1.
π =1.08
π3=
1.08
0.216= 5
β΄ πβ =5
1 β 0.6= 12.5
? a
? b
[Textbook] For a geometric series with first term π and common ratio π, π4 =15 and πβ = 16.a) Find the possible values of π.b) Given that all the terms in the series are positive, find the value of π.
π4 = 15 βπ 1 β π4
1 β π= 15 (1)
πβ = 16 βπ
1 β π= 16 (2)
Substituting π
1βπfor 16 in equation (1):
16 1 β π4 = 15
Solving: π = Β±1
2
As terms positive, π =1
2, β΄ π = 16 1 β
1
2= 8
Fro Warning: The power is π in the ππ formula but π β 1 in the π’π formula.
? a
? b
Test Your Understanding
Edexcel C2 May 2011 Q6
π =π
π
π = πππ
πΊβ = ππππ
πππ π βππ
π
π. ππ> ππππ
π >πππ
ππππ
πππ π. ππβ π = ππ
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Exercise 3E
Pearson Pure Mathematics Year 2/ASPages 75-76
Extension
[MAT 2006 1H] How many solutions does the equation2 = sin π₯ + sin2 π₯ + sin3 π₯ + sin4 π₯ + β―
have in the range 0 β€ π₯ < 2π
RHS is geometric series. π = π¬π’π§ π , π = π¬π’π§ π.
π =π¬π’π§ π
π β π¬π’π§ πβ π β ππ¬π’π§ π = π¬π’π§ π
π¬π’π§ π =π
πBy considering the graph of sin for π β€ π < ππ , this has 2 solutions.
1[MAT 2003 1F] Two players take turns to throw a fair six-sided die until one of them scores a six. What is the probability that the first player to throw the die is the first to score a six?
π· πππ πππ πππππ =π
π
π· πππ πππ πππππ =π
π
π
Γπ
π
π· πππ πππ πππππ =π
π
π
Γπ
π
The total probability of the 1st player winning is the sum of these up to infinity.
π =π
π, π =
π
π
π
=ππ
ππ
πΊβ =π/π
π β ππ/ππ=
π
ππ
2
[Frost] Determine the value of π₯ where:
π₯ =1
1+2
2+3
4+4
8+
5
16+
6
32+β―
(Hint: Use an approach similar to proof of geometric ππ formula)Doubling:
ππ = π +π
π+π
π+π
π+π
π+β―
Subtracting the two equations:
π = π +π
π+π
π+π
π+π
π+β―
= π + π = π
N
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Sigma Notation
π=1
5
(2π + 1)
What does each bit of this expression mean?
The Greek letter, capital sigma, means βsumβ.
The numbers top and bottom tells us what π varies between. It goes up by 1 each time.
We work out this expression for each value of π (between 1 and 5), and add them together.
= 3
π = 1
+5
π = 2
+7
π = 3
+9
π = 4
+11
π = 5
= 35If the expression being summed (in this case 2π + 1) is linear, we get an arithmetic series. We can therefore apply our usual approach of establishing π, π and π before applying the ππ formula.
Determining the value
π=1
7
3π
π=5
15
10 β 2π
First few terms? Values of π, π, π ππ π?
= 3 + 6 + 9 +β― π = 3, π = 3, π = 7
Final result?
π7 =7
26 + 6 Γ 3
= 84
= 0 + β2 + β4 +β― π = 0, π = β2, π = 11Be careful, there are 11 numbers between 5 and 15 inclusive. Subtract and +1.
π11 =11
20 + 10 Γ β2
= β110
π=1
12
5 Γ 3πβ1 = 5 + 15 + 45 +β― π = 5, π = 3, π = 12π12 =
5 1 β 312
1 β 3= 1328600
π=5
12
5 Γ 3πβ1 Note: You can either find from scratch by finding the first few terms (the first when π = 5), or by calculating:
π=1
12
5 Γ 3πβ1 β
π=1
4
5 Γ 3πβ1
i.e. We start with the first 12 terms, and subtract the first 4 terms.
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Solomon Paper A
Testing Your Understanding
Method 1: Direct
π=10
30
7 + 2π =27 + 29 + 31 +β―
π = 27, π = 2, π = 21
β΄ π21 =21
254 + 20 Γ 2
= 987
Method 2: Subtraction
π=10
30
7 + 2π =
π=1
30
7 + 2π β
π=1
9
7 + 2π
π=1
30
7 + 2π = 9 + 11 + 13 +β―
π = 9, π = 2, π = 30
π30 =30
218 + 29 Γ 2 = 1140
π9 =9
218 + 8 Γ 2 = 153
1140 β 153 = 987
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On your calculator
βUse of Technologyβ Monkey says:The Classwiz and Casio Silver calculator has a Ξ£ button.
Try and use it to find:
π=5
12
2 Γ 3π
Exercise 3F
Pearson Pure Mathematics Year 2/ASPages 77-78
Recurrence Relations
π’π = 2π2 + 3This is an example of a position-to-term sequence, because each term is based on the position π.
π’π+1 = 2π’π + 4π’1 = 3
But a term might be defined based on previous terms.If ππ refers to the current term, ππ+πrefers to the next term.So the example in words says βthe next term is twice the previous term + 4β
This is known as a term-to-term sequence, or more formally as a recurrence relation, as the sequence βrecursivelyβ refers to itself.
We need the first term because the recurrence relation alone is not enough to know what number the sequence starts at.
Example
Important Note: With recurrence relation questions, the the sequence will likely not be arithmetic nor geometric. So your previous π’π and ππ formulae do not apply.
Edexcel C1 May 2013 (R)
π₯2 = π₯12 β ππ₯1 = 1 β π
π =3
2
π₯3 = π₯22 β ππ₯2
= 1 β π 2 β π 1 β π= 1 β 3π + 2π2
= 1 + β1
2+ 1 + β
1
2+ β―
= 50 Γ 1 β1
2= 25
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?
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?Tip: When a Ξ£ question comes up for recurrence relations, it will most likely be some kind of repeating sequence. Just work out the sum of terms in each repeating bit, and how many times it repeats.
Test Your Understanding
Edexcel C1 Jan 2012
π₯2 = π + 5
π₯3 = π π + 5 + 5 = β―
π2 + 5π + 5 = 41π2 + 5π β 36 = 0π + 9 π β 4 = 0π = β9 ππ 4
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?
?
Combined Sequences
Sequences (or series) can be generated from a combination of both an arithmetic and a geometric sequence.
Example
Exercise 3G
Pearson Pure Mathematics Year 2/ASPages 80-81
[AEA 2011 Q3] A sequence {π’π} is given byπ’1 = ππ’2π = π’2πβ1 Γ π, π β₯ 1π’2π+1 = π’2π Γ π π β₯ 1
(a) Write down the first 6 terms in the sequence.
(b) Show that Οπ=12π π’π =
π 1+π 1β ππ π
1βππ
(c) [π₯] means the integer part of π₯, for example 2.73 = 2, 4 = 4.
Find Οπ=1β 6 Γ
4
3
π
2 Γ3
5
πβ1
2
[MAT 2014 1H] The function πΉ π is defined for all positive integers as follows: πΉ 1 = 0 and for all π β₯ 2,πΉ π = πΉ π β 1 + 2 if 2 divides π but 3 does not divide ,πΉ π = πΉ π β 1 + 3 if 3 divides π but 2 does not divide π,πΉ π = πΉ π β 1 + 4 if 2 and 3 both divide ππΉ π = πΉ π β 1 if neither 2 nor 3 divides π.Then the value of πΉ 6000 equals what?Solution: 11000
[MAT 2016 1G] The sequence π₯π , where π β₯ 0, is defined by π₯0 = 1 and
π₯π = Οπ=0πβ1 π₯π for π β₯ 1
Determine the value of the sum
π=0
β 1
π₯πSolution on next slide.
1
2 3
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?
Solution to Extension Question 3
[MAT 2016 1G] The sequence π₯π , where π β₯ 0, is defined by π₯0 = 1 and
π₯π = Οπ=0πβ1 π₯π for π β₯ 1
Determine the value of the sum
π=0
β 1
π₯π
Solution:ππ = πππ = πππ = π + π = πππ = π + π + π = π
This is a geometric sequence from ππ onwards.
Therefore
π=π
β π
ππ=π
π+π
π+π
π+π
π+π
π+β―
= π + π = π
A somewhat esoteric Futurama joke explained
This simplifies to
π = π0
π=1
β1
π
The sum 1
1+
1
2+
1
3+β― is known as the harmonic series, which is divergent.
Bender (the robot) manages to self-clone himself, where some excess is required to produce the duplicates (e.g. alcohol), but the duplicates are smaller versions of himself. These smaller clones also have the capacity to clone themselves. The Professor is worried that the total amount mass π consumed by the growing population is divergent, and hence theyβll consume to Earthβs entire resources.
Increasing, decreasing and periodic sequences
A sequence is strictly increasing if the terms are always increasing, i.e.π’π+1 > π’π for all π β β.e.g. 1, 2, 4, 8, 16, β¦
Similarly a sequence is strictly decreasing if π’π+1 < π’π for all π β β
A sequence is periodic if the terms repeat in a cycle. The order π of a sequence is how often it repeats, i.e. π’π+π = π’π for all π.e.g. 2, 3, 0, 2, 3, 0, 2, 3, 0, 2, β¦ is periodic and has order 3.
[Textbook] For each sequence:i) State whether the sequence is increasing, decreasing or periodic.ii) If the sequence is periodic, write down its order.
a) π’π+1 = π’π + 3, π’1 = 7 Terms: 7, 10, 13,β¦. Increasing.
b) π’π+1 = π’π2, π’1 =
1
2Terms:
1
2,1
4,1
8,1
16, β¦. Decreasing.
c) π’π+1 = sin 90πΒ° Terms: 1, 0, -1, 0, 1, 0, β¦ Periodic, order 4.
???
Textbook Error: It uses the term βincreasingβ when it means βstrictly increasingβ.
Exercise 3H
Pearson Pure Mathematics Year 2/ASPages 82-83
Modelling
Anything involving compound changes (e.g. bank interest) will form a geometric sequence, as there is a constant ratio between terms.We can therefore use formulae such as ππ to solve problems.
[Textbook] Bruce starts a new company. In year 1 his profits will be Β£20 000. He predicts his profits to increase by Β£5000 each year, so that his profits in year 2 are modelled to be Β£25 000, in year 3, Β£30 000 and so on. He predicts this will continue until he reaches annual profits of Β£100 000. He then models his annual profits to remain at Β£100 000.a) Calculate the profits for Bruceβs business in the first 20 years.b) State one reason why this may not be a suitable model.c) Bruceβs financial advisor says the yearly profits are likely to increase by 5% per annum.
Using this model, calculate the profits for Bruceβs business in the first 20 years.
The sequence is arithmetic up until Β£100 000, but stops increasing thereafter. We need to know the position of this term.
π = ππ πππ, π = ππππππ = π + π β π π β ππππππ= πππππ + π β π π β΄ π = ππ
First find sum of first 17 terms (where sequence is arithmetic):
πΊππ =ππ
π(π πππππ + ππ ππππ = π πππ πππ
β΄ πΊππ = π πππ πππ + π πππ πππ = π πππ πππ
? a
It is unlikely that Bruceβs profits will increase by exactly the same amount each year.
π = 20 000 π = 1.05
π20 =20000 1.0520 β 1
1.05 β 1= Β£ 661 319.08
? b
? c
a b
c
Geometric Modelling Example
[Textbook] A piece of A4 paper is folded in half repeatedly. The thickness of the A4 paper is 0.5 mm.(a) Work out the thickness of the paper after four folds.(b) Work out the thickness of the paper after 20 folds.(c) State one reason why this might be an unrealistic model.
π = 0.5 ππ, π = 2After 4 folds:π’5 = 0.5 Γ 24 = 8 ππ
After 20 foldsπ’21 = 0.5 Γ 220 = 524 288 ππ
It is impossible to fold the paper that many times so the model is unrealistic.
a
b
c
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Test Your Understanding
Edexcel C2 Jan 2013 Q3
? a
? b
? c
Exercise 3I
Pearson Pure Mathematics Year 2/ASPages 84-86
[AEA 2007 Q5]
The figure shows part of a sequence π1, π2, π3, β¦, of model snowflakes. The first term π1 consist of a single square of side π. To obtain π2, the middle third of each edge is replaced with a new
square, of side π
3, as shown. Subsequent terms are added by replacing
the middle third of each external edge of a new square formed in the
previous snowflake, by a square 1
3of the size, as illustrated by π3.
a) Deduce that to form π4, 36 new squares of side π
27must be added
to π3.
b) Show that the perimeters of π2 and π3 are 20π
3and
28π
3respectively.
c) Find the perimeter of ππ.
1
Extension
?
Just for your interestβ¦
Tell me about this βharmonic seriesββ¦
A harmonic series is the infinite summation:
1 +1
2+1
3+1
4+ β―
We can also use π»π to denote the sum up to the πth term (known as a partial sum), i.e.
π»π = 1 +1
2+1
3+ β―+
1
π
Is it convergent or divergent?The terms we are adding gradually get smaller, so we might wonder
if the series βconvergesβ to an upper limit (just like 1 +1
2+
1
4+
1
8+β―
approaches/converges to 2), or is infinitely large (i.e. βdivergesβ).
We can prove the series diverges by finding a value which is smaller, but that itself diverges:
π» = 1 +1
2+1
3+1
4+1
5+1
6+1
7+1
8+1
9+β―
π₯ = 1 +1
2+1
4+1
4+1
8+1
8+1
8+1
8+
1
16+ β―
Above we defined a new series π₯, such that we
use one 1
2, two
1
4π , four
1
8π , eight
1
16s and so on.
Each term is clearly less than (or equal to) each term in π». But it simplifies to:
π₯ = 1 +1
2+1
2+1
2+1
2+β―
which diverges. Since π₯ < π», π» must also diverge.
It is also possible to prove that π―π is never an integer for any π > π.
Where does the name come from?The βharmonic meanβ is used to find the average of rates. For example, if a cat runs to a tree at 10mph and back at 20mph, its average speed across
the whole journey is 131
3mph (not 15mph!).
Each term in the harmonic series is the harmonic mean of the two
adjacent terms, e.g. 1
3is the harmonic mean of
1
2and
1
4. The actual word
βharmonicβ comes from music/physics and is to do with sound waves.
10ππβ
20ππβ
The harmonic mean
of π₯ and π¦ is 2π₯π¦
π₯+π¦.
ππππππππππ
DrFrostMaths.com
The time Dr Frost solved a maths problem forMy mum, who worked at John Lewis, was selling London Olympics βtrading cardsβ, of which there were about 200 different cards to collect, and could be bought in packs. Her manager was curious how many cards you would have to buy on average before you collected them all. The problem was passed on to meβ¦
It is known as the βcoupon collectorβs problemβ.The key is dividing purchasing into separate stages where each stage involves getting the next unseen cardThere is an initial probability of 1 that a purchase results in an
unseen card. There is then a 199
200chance the next purchase
results in an unseen card. Reciprocating this probability gives
us the number of cards on average, i.e. 200
199, weβd have to buy
until we get our second unseen card. Eventually, there is a 1
200
chance of a purchase giving us the last unseen card, for which
weβd have to purchase 200
1= 200 cards on average. The total
number of cards on average we need to buy is therefore:200
200+200
199+ β―+
200
2+200
1
= 2001
200+
1
199+ β―+
1
2+ 1
= 200 Γ π»200 = 1176 πππππ
Youβre welcome John Lewis.
Really cool applications: How far can a stack of πbooks protrude over the edge of a table?
1 book: 1
2a length.
2 books: 1
2+
1
4=
3
4of a length.
1
2
3
4
3 books: 1
2+
1
4+
1
6=
11
12
11
12
Each of these additions is half the sum of the first π terms of the harmonic series.
e.g. 3 books: 1
21 +
1
2+
1
3=
11
12
The overhang for π books is therefore π
ππ―π
Because the harmonic series is divergent, we can keep adding books to get an arbitrarily large amount of overhang. But note that the harmonic series increases very slowly. A stack of 10,000 books would only overhang by 5 book lengths!