Transcript Year 9/GCSE: Sequences

Year 9: Sequences
Dr J Frost ([email protected])
Last modified: 1st October 2014
RECAP: Linear Sequences
4, 9, 14, 19, 24, 29, …
𝑛th term of sequence
The coefficient of 𝑛 is
the difference between
the numbers.
?
𝑢𝑛 = 5𝑛
−? 1
𝑛 means the position
in the sequence, so for
the first term, 𝑛 = 1.
5𝑛 alone would give us
the 5 times table. What
would we have to
correct it by to get the
desired sequence?
Quickfire Examples
Find formulae for the 𝑛th term of each of these sequences:
2, 5, 8, 11, 14, …
𝑢𝑛 = 3𝑛 −?1
6, 11, 16, 21, 26, …
𝑢𝑛 = 5𝑛 +?1
9, 7, 5, 3, 1, −1, …
𝑢𝑛 = 11 −?2𝑛
12, 9, 6, 3, 0, −3, …
𝑢𝑛 = 15 −?3𝑛
Terminology
 A linear sequence (or ‘arithmetic sequence’) is one where the
difference between terms is constant. 𝑢𝑛 = 𝑎𝑛 + 𝑏
 𝑢𝑛 means the 𝑛th term in the sequence.
(often 𝑎𝑛 or 𝑥𝑛 )
If we’re currently considering the
𝑛th term, how would we refer to the
term before that?
?
𝑈𝑛−1
This is the ‘position’.
𝑛
𝑢𝑛
1 2 3 4 5 6
𝟐 𝟓 𝟖 𝟏𝟏 𝟏𝟒 𝟏𝟕 …
This is the ‘term’.
i.e. The term at the 4th position is 11.
Exercise 1a
The 3rd term of a linear sequence is 17. The
4
1 Determine the formula for theth𝑛th term of each
45th term is 269. Determine the formula for
sequence. Hence find the 300 term for each sequence
the nth term.
(i.e. 𝑢300 )
𝒖𝒏 = 𝟔𝒏 − 𝟏
3, 8, 13, 18, …
5, 7, 9, 11, …
3, 4, 5, 6, …
5, 12, 19, 26, …
?
?
?
?
?
𝒖𝒏 = 𝟓𝒏 − 𝟐, 𝒖𝟑𝟎𝟎 = 𝟏𝟒𝟗𝟖
𝒖𝒏 = 𝟐𝒏 + 𝟑, 𝒖𝟑𝟎𝟎 = 𝟔𝟎𝟑
𝒖𝒏 = 𝒏 + 𝟐, 𝒖𝟑𝟎𝟎 = 𝟑𝟎𝟐
𝒖𝒏 = 𝟕𝒏 − 𝟐, 𝒖𝟑𝟎𝟎 = 𝟐𝟎𝟗𝟖
2 Find the first five terms of the sequences with the
following (non-linear) equations:
𝑢𝑛
𝑢𝑛
𝑢𝑛
𝑢𝑛
= 𝑛2 + 𝑛
= 𝑛3 − 𝑛2
= 3𝑛
= 𝑛!
?
?
?
?
𝒄𝒏 = 𝟐𝟏𝒏 + 𝟐
Whatever the first number is that coincides, we’ll see it 21
later because LCM(7,3)=21. Thus we know 𝒄𝒏 = 𝟐𝟏𝒏 + □.
It’s then simply a case of identifying which number this is (2).
?
at which they ‘grow’, the slowest growing first,
giving a reason for your order. It may help to
3 Find the formula for the 𝑛th term of the following
?
?
?
?
1 and 𝑏𝑛 = 7𝑛 + 2. A new sequence with
formula 𝑐𝑛 is formed by the numbers which
appear in both 𝑎𝑛 and 𝑏𝑛 . Determine 𝑐𝑛 .
N2 Put the sequences in order of the speed/rate
𝟐, 𝟔, 𝟏𝟐, 𝟐𝟎, 𝟑𝟎
𝟎, 𝟔, 𝟏𝟖, 𝟒𝟖, 𝟏𝟎𝟎
𝟑, 𝟗, 𝟐𝟕, 𝟖𝟏, 𝟐𝟒𝟑
𝟏, 𝟐, 𝟔, 𝟐𝟒, 𝟏𝟐𝟎
sequences.
6, 5, 4, 3, 2, …
𝒖𝒏 = 𝟕 − 𝒏
5, 2, −1, −4, …
𝒖𝒏 = 𝟖 − 𝟑𝒏
1
1
𝟓
10 , 8, 5 , 3, …
𝒖𝒏 = 𝟏𝟑 − 𝒏
2
2
𝟐
𝟏
𝟕
𝟓
𝟏
𝟏
𝟐𝟓
𝟐 ,𝟐
,𝟐 ,𝟑
𝒖𝒏 = 𝒏 +
𝟑 𝟏𝟐 𝟔 𝟏𝟐
𝟒
𝟏𝟐
N1 Two sequences have the formulae 𝑎𝑛 = 3𝑛 −
think what happens as 𝑛 increases by 1 each time.
𝑢𝑛 = 𝑛! 𝑢𝑛 = 𝑛2 𝑢𝑛 = 3𝑛
𝑛
𝑢𝑛 = 𝑛
𝑢𝑛 = 22 𝑢𝑛 = 𝑛
𝑢𝑛 = 1
1 is the slowest as it does not grow at all (1, 1, 1, 1, …). 𝒏 is next as the square root
causes the sequence to gradually grow slower over time. 𝒏 grows by 1 each time.𝒏𝟐
grows by 2𝑛 + 1 each time (as 𝑛 + 1 2 = 𝑛2 + 2𝑛 + 1) which means the difference
increases by a constant amount each time (in this case 2).𝟑𝒏 is next because the numbers
become 3 times larger each time, meaning the difference of the difference increases
unlike 𝑛2.𝒏! is next because the scale factor increases by 1 each time, whereas for 3𝑛 the
𝒏
scale factor is constant (i.e. 3).𝟐𝟐 is last because the scale factor doubles each time,
whereas for 𝑛! it only increased by 1.
?
Term-to-term and position-to-term formulae
1, 4, 7, 10, 13, 16, …
Position-to-term formula
We’ve previous seen how we can use the position 𝑛 to determine the term 𝑢𝑛 .
This is known as a position-to-term formula:
𝑢𝑛 = 3𝑛 − 2
Term-to-term formula
However, we can also get the 𝑛th term of the sequence by thinking of a rule to
get it from the previous term(s)…
Previous term
𝑢𝑛 = 𝑢𝑛−1 + 3
?
𝑢1 = 1
Why do you think we
need this?
Investigate
Describe these sequences.
Formula based on
previous terms
3, 5, 7, 9, …
𝑢𝑛 = 𝑢𝑛−1 + 2
?
𝑢1 = 3
Formula based on position 𝑛
𝑢𝑛 = 2𝑛 + 1
1
1, 1, 2, 3, 5, …
𝑢𝑛 = 𝑢𝑛−1 + 𝑢𝑛−2
?
𝑢1 = 1, 𝑢2 = 1
𝑢𝑛 =
2, 4, 8, 16, …
𝑢𝑛 = 2𝑢𝑛−1
?
𝑢1 = 2
𝑢𝑛 = 2𝑛
5
1+ 5
2
?
𝑛
1− 5
−
2
𝑛
Ermm…. ?
?
Bro Tip: This is an exam favourite!
Exercise 1b
For each of the following determine:
(a) the position-to-term formula and
(b) the term-to-term formula.
Sequence
1
2
3
4
5
6
N1
0, 1, 2, 3, 4, …
8, 6, 4, 2, …
3, 9, 27, 81, …
-1, +1, -1, +1, …
3, 6, 12, 24, …
0.5, 0.25, 0.125,
0.00625
0, 50, 75, 87.5, 93.75
P-to-T
T-to-T
𝑢𝑛 = 𝑛 − 1
?
𝑢𝑛 = 𝑢𝑛−1 + 1
𝑢1 = 0?
𝑢𝑛 = 10 − 2𝑛
?
𝑢𝑛 =
? 3𝑛
𝑢𝑛 = −1 𝑛
?
𝑢𝑛 = 𝑢𝑛−1 − 2
?
𝑢1 = 8
𝑢𝑛 = 3 × 2𝑛−1
𝑢𝑛 = 2𝑢𝑛−1
?
𝑢1 = 3
𝑢𝑛−1
𝑢𝑛 =
?2
𝑢1 = 0.5
𝑢𝑛−1
𝑢𝑛 =
?2 + 50
𝑢𝑛 =?3𝑢𝑛−1
𝑢𝑛 = −𝑢𝑛−1
𝑢1 = ?
−1
?
1
2?
𝑛
1
𝑢𝑛 = 100 −?100
2
𝑛−1
Year 9 Quadratic Sequences
Dr J Frost ([email protected])
Second Difference
What do you notice about the difference?
3, 8, 15, 24, 35, …
Working out 𝑢𝑛
𝑛
1
2 3
𝑢𝑛
3
8 15 24 35
+7
+5
+2
2
1𝑛
To add
1
2
4
5
+9
+2
4 9
4 6
16 25
8 10
2
𝑢𝑛 = 𝑛 + 2𝑛
STEP 1: Write
out 𝑛 and 𝑢𝑛
STEP 2: Work out
second difference.
STEP 3: Halve this
to find coefficient
of 𝑛2 term.
STEP 4: Work out
what we need to add
to get from this to
correct term. Work
out its formula.
More Examples
2
3, 9, 19, 33, 51, …
𝑢𝑛 = 2𝑛 + 1?
4, 7, 11, 16, 22, …
1 2 3
𝑢𝑛 = 𝑛 + ? 𝑛 + 2
2
2
4, 15, 32, 55, 84, …
? −1
𝑢𝑛 = 3𝑛2 + 2𝑛
4, 9, 18, 31, 48, …
2
? +3
𝑢𝑛 = 2𝑛 − 𝑛
Exercise 2
4
3, 7, 13, 21, 31, …
5, 13, 25, 41, 61, 85, …
2, 3, 6, 11, 18, …
5, 17, 35, 59, 89, 125, …
𝑢𝑛
𝑢𝑛
𝑢𝑛
𝑢𝑛
5
1.5, 4, 7.5, 12, 17.5, 24, …
𝑢𝑛
1, 10, 23, 40, 61, 86, …
4, 5, 4, 1, −4, −11, …
𝑢𝑛
𝑢𝑛
3.5, 7, 9.5, 11, 11.5, 11, …
𝑢𝑛
1
2
3
6
7
8
= 𝑛2 + 𝑛?+ 1
? +1
= 2𝑛2 + 2𝑛
? +3
= 𝑛2 − 2𝑛
? −1
= 3𝑛2 + 3𝑛
1 2
= 𝑛 + ?𝑛
2
= 2𝑛2 + 3𝑛
? −4
= −𝑛2 + ?4𝑛 + 1
1 2
= 𝑛 + ?5𝑛 − 1
2
Levelled Sequences Activity
Work alone or in pairs.
Complete all Level 1 questions before checking your
answers.
If correct, you can proceed to Level 2 and so on.
Year 9 Further Sequences
Dr J Frost ([email protected])
The general form of...
A linear (“first difference”) sequence:
𝑢𝑛 = 𝑎𝑛 +?𝑏
A quadratic (“second difference”) sequence:
? +𝑐
𝑢𝑛 = 𝑎𝑛2 + 𝑏𝑛
Why does the first difference...
...become the number on front of the n?
Current Term
Position
Term
Next Term
𝑛
𝑎𝑛 + 𝑏
𝑛 +? 1
𝑎𝑛 + ?𝑎 + 𝑏
a?
Why does the second difference...
...get halved then put on front of n2?
Current Term
Position
Term
Next Term
Term after that
𝑛
? 1
𝑛+
𝑛 +? 2
𝑎𝑛2 + 𝑏𝑛 + 𝑐
𝑎𝑛2 + 2𝑎𝑛
? + 𝑏𝑛
+𝑎+𝑏+𝑐
2𝑎𝑛 +?𝑎 + 𝑏
𝑎𝑛2 + 4𝑎𝑛 + 𝑏𝑛
? +𝑐
+ 4𝑎 + 2𝑏
2𝑎𝑛 + ?3𝑎 + 𝑏
?
2𝑎
Since the second difference is 2𝑎 and the coefficient of 𝑛2 is 𝑎, we can
see halving the second difference gives us the coefficient of 𝑛2 .
Finding a formula using simultaneous equations
You’re given the first three terms of a quadratic (second difference) sequence:
𝒖𝟏 = 𝟑,
We know that we can use:
𝒖𝟐 = 𝟕,
𝒖𝟑 = 𝟏𝟓
𝒖𝒏 = 𝒂𝒏𝟐 + 𝒃𝒏 + 𝒄
What equations can we form?
𝒂+𝒃+𝒄
=𝟑
𝟒𝒂 + 𝟐𝒃 + 𝒄 =?𝟕
𝟗𝒂 + 𝟑𝒃 + 𝒄 = 𝟏𝟓
(𝟏)
(𝟐)
(𝟑)
Solve by elimination:
2 − 1 : 3𝑎 + 𝑏 = 4
3 − 1 : 8𝑎 + 2𝑏 = 12
?
𝑎 = 2, 𝑏 = −2, 𝑐 = 3
𝑢𝑛 = 2𝑛2 − 2𝑛 + 3
Test Your Understanding
1, 4, 13, ...
2
? +4
𝑢𝑛 = 3𝑛 − 6𝑛
Oxford Maths Admissions Exam - 2009
x4 = 10, x5 ?
= 15
A = 0, B = 0.5,
? C = 0.5
n = 40
?
Further Exercises
Solve the following by forming simultaneous equations.
1
Given that 𝑢1 = 2, 𝑢2 = 7, 𝑢3 = 14 and that the formula for the sequence is
𝑢𝑛 = 𝑎𝑛2 + 𝑏𝑛 + 𝑐, form simultaneous equations, and hence determine 𝑎, 𝑏 and 𝑐.
𝒂 = 𝟏, 𝒃 = 𝟐, 𝒄 = −𝟏
?
2
A line with equation 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 goes through the points 1,0 , 2, 7 , 3, 18 .
Determine 𝑎, 𝑏 and 𝑐.
𝒂 = 𝟐, 𝒃 = 𝟏, 𝒄 = −𝟑
?
3
A line with equation 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 goes through the points 2,10 , (4,46) and 5,73 .
Determine 𝑎, 𝑏 and 𝑐.
𝒂 = 𝟑, 𝒃 = 𝟎, 𝒄 = −𝟐
?
2
4
A line has equation 𝑦 = 𝑥𝑎 𝑥 , where 𝑎 is a constant. It passes through the point 2, 162 .
Determine the 𝑦-value when 𝑥 = 3.
𝒚 = 𝟑𝟏𝟎 = 𝟓𝟗𝟎𝟒𝟗
By forming suitable simultaneous equations (or otherwise), determine the formula for the nth
term of the sequence…
3
1, 1, 1, 2, 5, 11, 21, …
Prove
that
the
coefficient
of
the
𝑛
term in a
NN
1
𝟏
𝟏𝟏
cubic sequence is of the third difference.
𝒖𝒏 = 𝒏𝟑 − 𝒏𝟐 +
𝒏
6
𝟔
𝟔
More generally, if the 𝑘 th difference was
constant, what do you think the coefficient of
the 𝑛𝑘 term will be? (NNN Prove it.)
?
N
?