seq and series notes

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Transcript seq and series notes

Sequence
A sequence is a set of numbers
in a specific order
Infinite sequence
Finite sequence
a1 , a2 , a3 , a4 ,..., an ,...
a1 , a2 , a3 , a4 ,..., an
Sequences –
sets of numbers
Notation:
an   represents the formula for finding terms 
n  term number
a4 is the notation for the 4th term
a32 is the notation for the 32nd term
Examples:
If an  2n  3, find the first 5 terms.
If an  3n  1, find the 20th term.
.
Ex 1
Find the first four terms of the sequence
an  3n  2
a1  3(1)  2  1
First term
a2  4
Second term
a3  7
Third term
a4  10
Fourth term
Writing Rules for Sequences
We can calculate as many terms as we want as
long as we know the rule or equation for an.
Example:
3, 5, 7, 9, ___ , ___,……. _____ .
an = 2n + 1
Series –
the sum of a certain
number of terms of a sequence
n
Sigma Notation :
a
i 1
Stop
Formula
i
Start
“Add up the terms in the sequence
beginning at term number 1 and going
through term number “n”.
4
1.  -5i  5 1  5  2  5  3  5  4  50
i 1
5
2. 
i 1
7
1
2
3.  i
i3
6
4.
3
i 1
25
i 1 
2
 25
Infinite Sequence
a1 , a2 , a3 , a4 ,..., ai ,...

Infinite Series
a1  a2  a3  a4  ...  ai  ...   ai
i 1
Finite Series or nth Partial Sum
n
a1  a2  a3  a4  ...  an   ai
i 1
SUMMATION NOTATION
Sum of the terms of a finite sequence
Upper limit of summation
(Ending point)
n
a
i 1
i

Lower limit of summation
(Starting point)
Sequence
• There are 2 types of Sequences
Arithmetic:
You add a common difference each
time.
Geometric:
You multiply a common ratio each time.
Arithmetic Sequences
Example:
• {2, 5, 8, 11, 14, ...}
Add 3 each time
• {0, 4, 8, 12, 16, ...}
Add 4 each time
• {2, -1, -4, -7, -10, ...}
Add –3 each time
Arithmetic Sequences
• Find the 7th term of the sequence:
2,5,8,…
Determine the pattern:
Add 3 (known as the common difference)
Write the new sequence:
2,5,8,11,14,17,20
So the 7th number, or a7, is 20
Arithmetic Sequences
• When you want to find a large sequence,
this process is long and there is great room
for error.
• To find the 20th, 45th, etc. term use the
following formula:
an = a1 + (n - 1)d
Arithmetic Sequences
an = a1 + (n - 1)d
Where:
a1 is the first number in the sequence
n is the number of the term you are
looking for
d is the common difference
an is the value of the term you are
looking for
Arithmetic Sequences
• Find the 15th term of the sequence:
34, 23, 12,…
Using the formula an = a1 + (n - 1)d,
a1 = 34
d = -11
n = 15
an = 34 + (n-1)(-11) = -11n + 45
a15 = -11(15) + 45
a15 = -120
Arithmetic Sequences
Melanie is starting to train for a swim meet.
She begins by swimming 5 laps per day for
a week. Each week she plans to increase her
number of daily laps by 2. How many laps
per day will she swim during the 15th week
of training?
Arithmetic Sequences
• What do you know?
an = a1 + (n - 1)d
a1 = 5
d= 2
n= 15
t15 = ?
Arithmetic Sequences
• tn = t1 + (n - 1)d
• tn = 5 + (n - 1)2
• tn = 2n + 3
• t15 = 2(15) + 3
• t15 = 33
During the 15th week she will swim 33 laps per
day.
Arithmetic Series
• The sum of the terms in a sequence are
called a series.
• There are two methods used to find
arithmetic series:
Formula
Sigma Notation
Geometric Sequences
•
In geometric sequences, you multiply by a
common ratio each time.
•
1, 2, 4, 8, 16, ...
multiply by 2
27, 9, 3, 1, 1/3, ...
Divide by 3 which means multiply by 1/3
•
Geometric Sequences
• Find the 8th term of the sequence:
2,6,18,…
Determine the pattern:
Multiply by 3 (known as the common ratio)
Write the new sequence:
2,6,18,54,162,486,1458,4374
So the 8th term is 4374.
Geometric Sequences
• Again, use a formula to find large numbers.
•
an = a1 • (r)n-1
Geometric Sequences
• Find the 10th term of the sequence :
4,8,16,…
an = a1 • (r)n-1
• a1 = 4
• r=2
• n = 10
Geometric Sequences
an = a1 • (r)n-1
a10 = 4 • (2)10-1
a10 = 4 • (2)9
a10 = 4 • 512
a10 = 2048
Geometric Sequences
• Find the ninth term of a sequence if
a3 = 63 and r = -3
a1= ?
n= 9
r = -3
a9 = ?
There are 2 unknowns so you must…
Geometric Sequences
• First find t1.
• Use the sequences formula substituting t3 in for
tn. a3 = 63
• a3 = a1 • (-3)3-1
• 63 = a1 • (-3)2
• 63= a1 • 9
• 7 = a1