Unit 7: Sequences + Series
Day 1: Intro to Sequences/Series
Definitions:
Sequence:
A set of numbers in order that follow a predictable pattern.
Terms:
Each number in the sequence is called a term
tn:
Refers to the nth term in a sequence or series. For example, t10 is the 10th term.
Series:
The sum of a sequence of numbers
Sn :
The sum of the first n terms in a series. For example, S10 is the sum of the first 10 terms.
Recursion:
When each term’s value depends of previous terms.
Sequence or Series
2, 4, 6, 8, 10, …
1 + 3 + 5 + 7 +…
1, 1, 2, 3, 5, 8, …
1+2+4+8+…
240, 120, 60, …
Sequence
Series
Sequence
Series
Sequence
Series
Sequence
Series
Sequence
Series
Next 3 terms
S5
12
30
9
13
16
30
14
11
21
32
15
16
13
34
64
7.5
Equation of tn
Explicit
tn = 2 + 2(n – 1)
Recursive
t1 = 2; tn = tn-1 + 2
Explicit
tn = 1 + 2(n – 1)
Recursive
t1 = 1; tn = tn-1 + 2
Explicit
We don’t study this type of sequence
Recursive
t1 = 1; t2 = 1; tn = tn-1 + tn-2
Explicit
tn = 1 ∙ 2
Recursive
t1 = 1; tn = 2(tn-1)
Explicit
tn = 240 ∙
Recursive
t1 = 240; tn = (tn-1)
25
12
31
465
For the sequence below, choose any value you like for t1 and then write as many terms as you can
Starting with t1 = 50:
50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10,
5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1 …
Starting with t1 = 99:
99, 298, 149, 448, 224, 112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40,
20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1 …
Nobody knows if there is a t1 for which this sequence doesn’t end with the repetition of “4, 2, 1…”
tn =
{
3tn-1 + 1 if tn-1 is odd
if tn-1 is even
Day 2: Arithmetic Sequences
2, 6, 10, 14, 18, 22, …
What are the next three terms in the sequence?
Example
: What is the formula for tn?
There are two parts to the formula: The first term (called a) and the common difference between terms (d)
Each term in the sequence can be written as a combination of a and d
2
6
10
14
18
2
a
2+4
a+d
2+4+4
a + 2d
2+4+4+4
a + 3d
2+4+4+4+4
a + 4d
Based on this pattern, tn = a + (n – 1)d
Example
: Answer the questions below about the arithmetic sequence
21, 17, 13, 9, 5, …
a.) Write an explicit and a recursive formula to define
the sequence
Explicit
tn = 21 – 4(n – 1)
Recursive
t1 = 21; tn = tn-1 – 4
b.) Find the 12th term, and the 650th term.
t12 = 21 – 4(12 – 1) = -23
t650 = 21 – 4(650 – 1) = -2575
c.) Is the value -999 part of the sequence?
d.) Is the value -400 part of the sequence?
If so, -999 = 21 – 4(n – 1) for some integer n.
If so, -400 = 21 – 4(n – 1) for some integer n.
n = 256, so -999 is t256
n = 106.25, so -400 is not in the sequence.
(However, we know -400 must lie between the values of t106 and t107)
Example
: Find the explicit and recursive equation of an arithmetic sequence where t3 = 11, and t9 = 29
1. Goal: Find “a” and “d”
2. t3 and t9 are 6 terms apart so their values
are 6 common differences apart.
3. 29 – 11 = 6d gives us d = 3
4. t9 = 29 = a + (n – 1)d
29 = a + (9 – 1)(3)
29 = a + 24
5=a
5. Equations:
Explicit: tn = 5 + (n – 1)(3)
Recursive: t1 = 5; tn = tn-1 + 3
Day 3: Geometric Sequences
2, 6, 18, 54, 162, …
What are the next three terms in the sequence? 486, 1458, 4374, … (Multiplying by 3 each time)
Example
: What is the formula for tn?
There are two parts to the formula: The first term (called a) and the common ratio between terms (r)
Each term in the sequence can be written as a combination of a and r
2
6
18
54
162
2
a
2×3
ar
2×3×3
ar2
2×3×3×3
ar3
2×3×3×3×3
ar4
Based on this pattern, tn = ar(n – 1)
The sequence above has the formulas below
Explicit:
= (2) ∙ (3)
Recursive:
= 2;
=3∙
Example
: Answer the questions below about the geometric sequence
–, , –, 1, –2, 4, –8, …
a.) Write an explicit and a recursive formula to define
the sequence.
b.) Find the 12th term, and the 20th term.
1
= − ∙ (−2)
8
Starts at − , each term is -2 times the last
= − ;
Recursive:
= −2 ∙
1
= − ∙ (−2)
8
= − ∙ (−2)
Explicit:
= 256
= 65536
c.) Is the value 1048576 part of the sequence?
d.) What is the first term greater than 1 billion?
If so, then 1048576 = − ∙ (−2)
Find the position of 1 billion in the list
1
1000000000 = − ∙ (−2)
8
for an integer n
1
1048576 = − ∙ (−2)
8
−8388608 = (−2)
−8000000000 = (−2)
Ignore signs, Guess & Check
2
Ignore signs, Guess & Check
(−2)
= (−2)
∴ 1048576 is t24
= (2)
32.9 = n – 1
23 = n – 1
24 = n
.!
33.9 = n
∴ The 33 term has magnitude under 1 billion
rd
The 34th term has magnitude over 1 billion
If t34 is negative, t35 will be the first term over 1 billion.
Check:
"
= − ∙ (−2)
"
= 1073741824 > 1000000000
∴The 34th term is the first term over 1 billion
Example
: Find the explicit and recursive equation of a geometric sequence where t3 = 250, and t8 = 781250
We have to find a, and find r
Find r
Since the given terms are 5 apart, t8 is t3 multiplied by r 5 times
∴ t8 = t 3 x r 5
781250 = 250 r5
3125 = r5
$
√3125 = %
5=r
Find a
=&∙%
250 = & ∙ 5
250 = & ∙ 25
10 = &
Equations
Starts at 10, each term is 5 times the last
Recursive:
Explicit:
= 10;
=5∙
= 10 ∙ (5)
Day 4: Arithmetic Series
Sum of the first n terms
2 + 6 + 10 + 14 + 18 + 22 + …
Example
Sn =
[2a + (n – 1)d]
: Find each sum as indicated using the formula:
S1
S2
S3
S4
1
' = (2(2) + (1 − 1)(4)*
2
2
' = (2(2) + (2 − 1)(4)*
2
3
' = (2(2) + (3 − 1)(4)*
2
4
'" = (2(2) + (4 − 1)(4)*
2
2
8
18
32
Note: The terms of an arithmetic sequence grow at a constant rate. A consequence of this is we can only add up a finite number of
terms. With some geometric sequences it will be possible to add up an infinitely long list of numbers, depending on the rate of growth.
Example
: Derivation of the formula for Sn:
omit
Example
: Find the sum of the first 25 terms of the sequence 2 + 5 + 8 + 11 + …
The series is arithmetic since each term is 3 more than the previous term
n = 25; a = 2; d = 3
'
+
=
25
(2(2) + (25 − 1)(3)*
2
'
Example
+
= 950
: Find the sum of the finite sequence 99, 91, 83, 75, … -117
The series is arithmetic since each term is 8 less than the previous term
n = ?; a = 99 d = –8
Find S28
Find n
tn = a + (n – 1)(d)
–117 = 99 + (n – 1)(–8)
–216 = (n – 1)(–8)
27 = (n – 1)
28 = n
'
=
28
(2(99) + (28 − 1)(−8)*
2
' = −252
Day 5: Geometric Series
Sum of the first n terms
3 + 6 + 12 + 24 + 48 + 96 + …
Example
Sn =
: Find each sum as indicated using the formula:
-(. /
)
.
a = 3; r = 2
S1
' =
3(2 − 1)
2−1
=1
S2
' =
S3
3(2 − 1)
2−1
' =
=9
3(2 − 1)
2−1
= 21
S4
'" =
3(2" − 1)
2−1
= 45
Notes: See the end of the solutions for a discussion of these notes.
•
The formula above doesn’t work if r = 1. What would a geometric sequence with r = 1 look like? Can you come up with a
formula for Sn that works for a geometric series with r = 1?
•
It is impossible to add up an infinite geometric series if r > 1 because the terms grow too fast.
•
It is sometimes possible to add up an infinite geometric series if r is a fraction between 0 and 1. A different formula is needed.
Example
: Derivation of the formula for Sn:
omit
Example : Find the sum of the first 25 terms of the
sequence 3 + 6 + 12 + 24 + 48 + 96 + …
n = 25;
a = 3;
S25 =
Example
Example : Find the sum of the first 25 terms of the
sequence 3 – 6 + 12 – 24 + 48 – 96 + …
r=2
( 0$
n = 25;
)
S25 =
a = 3;
r = –2
)0$
((
= 100663293
= 33554433
Over 100 million
Only 33 million
: Find the sum of 27 + 18 + 12 + … +
+1
"
+
)
+
2 !
The terms start at 27 and decrease by a factor of  each time.
n = ?;
Find n
a = 27;
r=
Find S10
2
= (27) ∙ 3 4
3
2
512
= (27) ∙ 3 4
3
729
2
512
=3 4
3
19683
Guess & Check
2 !
2
3 4 =3 4
3
3
9=n–1
10 = n
S10 =
=
25
2:
2
67
3
0
9
1024
59049
6
9
8
;
= 79.59533608
Notes: Geometric Series Discussion (Non-Testable)
•
The formula above doesn’t work if r = 1. What would a geometric sequence with r = 1 look like? Can you come up with a
formula for Sn that works for a geometric series with r = 1?
A geometric series with r = 1 would look like this:
a+a+a+a+a+…
Where each term is equal to the first term. Therefore the sum would be Sn = (a)(n)
•
It is impossible to add up an infinite geometric series if r > 1 because the terms grow too fast.
For example, 1 + 2 + 4 + 8 + 16 + …
The more numbers you add, the bigger the sum gets. It keeps getting bigger and
bigger.
S1 = 1
S2 = 1 + 2
S3 = 1 + 2 + 4
S4 = 1 + 2 + 4 + 8
S5 = 1 + 2 + 4 + 8 + 16
…
S25 = 1 + 2 + 4 + … + 16777216
=1
=3
=7
= 15
= 31
= 33554431
•
It is sometimes possible to add up an infinite geometric series if r is a fraction between 0 and 1. A different formula is needed.
For example, with the series + + + +
"
6 /
Sn =
'
'
'
'"
'+
'1
…
'
5 0
6
0
1
+⋯
we have & = 1; %
8
only if n is finite. The sum formula simplifies to
'
/
/ 6
=1
)
) )
= 1.5
"
= 1.75
"
= 1.875
) ) )
) ) ) )
"
) )") )
1
1
) ) ) ⋯)
"
= 1.9375
)
" +21
= 1.96875
= 1.999998093
The more terms you add, the bigger the sum gets. It keeps getting bigger and bigger but seems to level off
towards a total of 2. In fact if you add up an infinite number of terms you will get a sum of exactly 2 for this
sequence.
Other weird things can happen when you add up an infinite number of fractions…
•
Alternating the signs of reciprocals of powers of 2
•
The reciprocals of square numbers
•
The reciprocals of triangular numbers
•
Alternating the signs of the reciprocals of odd integers
You do this kind of stuff if you study sequences/series in university.