Further Exploration
of Patterns
Abstract Quadratic Patterns
a+b+c
1st change
4a+2b+c
3a+b
2nd change
9a+3b+c
5a+b
2a
16a+4b+c
7a+b
2a
25a+5b+c
9a+b
2a
Coefficient of n2 is
1
(2a)  a.
2
Constant second change, therefore it is quadratic starting with an2.
Write out the an2 series a, 4a, 9a, 16a,... and subtract from the original series.
a  b  c, 4a  2b  c, 9a  3b  c, 16a  4b  c, 25a  5b  c,

a
4a
9a
16a
25a,
bc
2b  c
3b  c
4b  c
5b  c,
1st change
b
b
b
b
This is linear bn
If you now go down form the 1st term which is (b  c)  b  c
The General Term is now an2  bn  c
Generalising Quadratic Sequences
n2
1,
4,
3
1st change
2
8,
6
1st change
10
4
4n2
16,
4,
0.5n2
1st change
2nd change
14
18
4
36,
20
8
0.5,
9
2
4
12
2nd change
25,…
18, 32, 50,…
2nd change
1st change
7
2
2,
16,
5
2nd change
2n2
9,
2,
1.5
64,
28
8
36
8
4.5,
2.5
1
100,...
8,
3.5
1
12.5,...
4.5
1
When you get half the 2nd change
you get the coefficient of n2
Your Turn [2.7 pg. 18]
Write the General Formula for the following paterns.
(a)
3,
12,
(b)
0.25,
27,
1,
2.25,
Solutions:
(a)
3n2
(b) 0.25n2
48,
75,…
4,
6.25,…
More Difficult Quadratic Patterns – Method 1
6,
1st change
2nd change
12,
3
20,
5
2
30,
7
2
42,…
9
2
Therefore it is n2 with other terms.
Write out the n2 series 1, 4, 9, 16,... and subtract from the original series.
6, 12, 20, 30, 42,
 1, 4, 9, 16, 25,
5, 8, 11, 14, 17,
1st change 3
3
3
3
This is linear 3n
If you now go down form the 1st term which is 5  3  2
The general term is now n2  3 n  2
More Difficult Quadratic Patterns – Method 2
6, 12, 20, 30, 42,...
Look for lowest common factor, here it is 2
2  3, 3  4, 4  5, 5  6, 6  7,...
This is an AP  AP
First AP
2, 3, 4, 5, 6,.. Tn  a  (n  1)d  2  (n  1)1  n  1
Second AP
3, 4, 5, 6, 7,... Tn  a  (n  1)d  3  (n  1)1  n  2
Tn  Tn  (n  1)(n  2)  n2  3n  2
Your Turn [2.8 pg. 18]
Write the General Formula for the following paterns.
(a)
5,
12,
21,
32,
45, ..........................
(b)
5,
15,
31,
53,
81, ..........................
Solutions:
(a) n2 + 4n
(b) 3n2 + n + 1
Your Turn [2.9 pg. 18]
1st
2nd
3rd
How many blocks are in the 4th pattern?
Write a gerneral formula to find the number of in the nth pattern.
How many blocks are in the 8th pattern?
SOLUTION
1st
4
1st change
2nd change
3rd
2nd
12
8
24
12
4
40
16
4
Constant, therefore quadratic
4th
Method 1
4
1st change
2nd change
12
8
24
12
4
40
16
4
2
Therefore it is 2n2 with other terms.
Write out the 2n2 series 2, 8, 18, 32,... and subtract from the original series.
4, 12, 24, 40,...
 2, 8, 18, 32,...
2, 4, 6, 8,
2 2
2
This is linear 3n
1st change
If you now go down form the 1st term which is 2  2  0
The general term is now 2n2  2 n
T8  2(8)2  2(8)  144 blocks
Method 2
4, 12, 24, 40,
Look for lowest common factor, here it is 2
[Looking at length  breadth]
2  2, 3  4, 4  6, 5  8,
This is an AP  AP
First AP
2, 3, 4, 5,...
Second AP
3, 4, 5, 6, 7,... Tn  a  (n  1)d  2  (n  1)2  2n
Tn  Tn  (n  1)(2n)  2n2  2n
T8  2(8)2  2(8)  144 blocks
Tn  a  (n  1)d  2  (n  1)1  n  1
Graphing the Couples
T8 = 144
Given you have 20 metres of wire, what is the maximum rectangular shaped
area that you can enclose?
1
2
9
1st change
2nd change
4
6
7
8
9
Area
3
16
7
21
5
–2
24
3
–2
Therefore it is  n2 with other terms.
Write out the  n2 series  1,  4,  9,  16,... and subtract from the original series.
9, 16, 21, 24,...
 1, 4, 9, 16,...
This is linear 10n
10, 20, 30, 40,
The general term for the area is  n2  10 n
Geometric
T1  2
T2  4
T3  8
T4  16
Tn  ?
This is a Geometric Sequence with T1  2  a
Each term is multiplied by 2 to get the next term: r(Common Ratio)  2
Tn
r
Tn1
Finding the Tn of a Geometric Sequence
T1  a
T2  ar
T3  (ar )r  ar 2
T4  (ar 2 )r  ar 3
T5  (ar 3 )r  ar 4
Tn  ar n1
Exponential
A ball is dropped from a height of 8 m. The ball bounces to 80% of its previous
height with each bounce. How high (to the nearest cm) does the ball bounce
on the fifth bounce.
8,
1st
2nd
6.4,
5.12,
3rd
4.96,
4th
3.2768,
5th
2.62144
2.62144m = 262 cm
Is this a Linear, Quadratic or Cubic pattern?
Let’s look at the graph of this.
Blue:
f(x)  8(0.8)x
or
Tn  8(0.8)n
Write down the function which describes the red graph.
What is the total distance travelled by the ball when it hits the ground for
the 5th time?
What if we were asked to find the total distance travelled when the ball
hits the ground for the 20th time. Is there any general way of doing it?
a(r n  1)
Sn 
r 1
a(1 r n )
Sn 
1 r
Finding the Sn of a Geometric Series
Sn
 a  ar  ar 2  ar 3  ar 4 
rSn

Sn  rSn  a
ar  ar 2  ar 3  ar 4 
0


0

0
 ar n1
 ar n1  ar n

0

 ar n
Subtracting
(1 r )Sn  a  ar n
a  ar n
Sn 
1 r
a(1 r n )
a(r n  1)
Sn 
or Sn 
1 r
r 1
These formulas can also be proved by Induction
Link to Student’s CD
A rabbit is 10 metres away from a some food. It hops 5 metres, then hops
2.5 metres, then 1.25 metres, and so on, hopping half its previous hop each
time. What will the length of the 6th hop be?
5m
2.5 m
1.25 m
5,
2.5,
1.25,
0.625,
0.3125,…
What type of pattern is this? Discuss
A rabbit is 10 metres away from some food. It hops 5 metres, then hops
2.5 metres, then 1.25 metres, and so on, hopping half its previous hop each
time.
If the rabbit kept hopping forever, what in theory would be the total distance
travelled by it?
5m
2.5 m
1.25 m
The Sum to Infinity of a GP
For r  1
r  0
a(1 r n )
a
ar n
Sn 


1 r
1 r 1 r
a
0
S 

as r   0 for r  1
1 r 1 r
S 
a
for r  1
1 r
Extending the Blocks Question
Find the total number of blocks required to make the first 25 patterns.
Tn  2n2  2n
n
n
Sn  2 r  2 r
r 1
2
r 1
 n(n  1)(2n  1)
 n(n  1)
Sn  2 

2

 2 
6




 n(n  1)(2n  1)
Sn  
 n(n  1)

3


 25(25  1)(2(25)  1)
S25  
 25(25  1)

3


S25  11,700
Three Formulae
n(n  1)
r


2
r 1
n
n(n  1)(2n  1)
2
r


6
r 1
n
 n(n  1) 
3
r


 2 


r 1
n
These formulas can be proved by Induction
2
Summary of GP formulae
Tn of a GP  ar n1
Sn
a(r n  1)
of an GP 
for r  1
r 1
or
Tn  Sn  Sn1
a
S 
1 r
for r  1
a(1 r n )
for r  1
1 r
Express 1.2 in the form of
a
where a and b 
b
Method 1
1.2  1.222222222
 1 0  2  0  02  0  002  0  0002  0  00002 
2
2
2
2
 1  



10
100
1000
10000

This is an infinite GP with a 
S 
S 
a
1 r
2
10
1 101
1.2  1

2
9
2 11

9 9



2
1
and r 
10
10
Method 2
Let x  1.222222222
10x  12.22222222
x  1.222222222
9x  11
x

11
9
Express 1.43 in the form of
a
where a and b 
b
Method 1
1.43  1.43434343
 1 0  43  0  0043  0  000043 
Method 2
Let x  1.43434343
100x  143.43434343


x  1.4343434343

43
1 99x  142
This is an infinite GP with a 
and r 
100
100
142
x 
a
99
S 
1 r
43
43
100
S 

1
1 100
99
43
43
 43
 1 



100
10000
1000000

1.2  1
43 142

99
9
Other Types of Series
1
AP  AP
1
1
1


n(n  1)
n n 1
Fill in the various values in the square brackets and find the sum to n terms of the series
1
whose Tn is
n(n  1)
Find the sum of the first 20 terms of the series
1
1
T1 

   
Show that
1
1

   
1
1
T3 

   
T2 
1
1

   
1
1
Tn 

   
Tn1 
Sn 
Solution
1 1
T1  
1 2
1 1
T2  
2 3
1 1
T3  
3 4
1
1

n 1 n
1
1
Tn  
n n 1
1
Sn  1
n 1
1
S20  1
20  1
20
S20 
21
Tn1
Arithmethico Geometric AP x GP
1 2r  3r 2 
 nr(n1)
Find the Tn of the following sequence 2, 8, 24, 64, 160
Find the Tn of the following sequence 1 2, 2  4, 3  8, 4  16, 5  32
Each term in this sequence is an AP  GP
Tn of GP  2(2)n1  2 n
Tn of AP  n
Combined, Tn  n(2 n )
GeoGebra Three Graphs
Function Inspector
2012 LCHL Q4
In a science experiment, a quantity Q(t) was observed at various points in
time t. Time is measured in seconds from the instant of the first observation.
The table below gives the results.
t
0
1
2
3
4
Q(t)
2920
2642
2391
2163
1957
Q follows the rule of the form Q(t)  Aebt , where A and b are constants.
(a) Use any two of the observations from the table to find the value
of A and the value of b, correct to three decimal places.
(b) Use a different observation from the table to verify your values for
A and b.