Transcript 6 -6 Factoring by Grouping

Write the next three numbers in each pattern.
1) 15,11,7,3,-1,…
2) -12,-7,-2,3,8,…
3) 10,11,13,16,20,…
4) 2,6,18,54,162,…
5) 96,48,24,12,6,…
6) 1,4,9,16,25,…
Sequences and Series
Sequences
A sequence is an ordered list of numbers, called
terms.
An infinite sequence continues without end while a
finite sequence has a last term.
A formula that defines the nth term of a sequence is
called an explicit formula.
A recursive formula is one where one or more
previous terms are used to generate the next term.
Example 1
Write the first six terms of the sequence
defined by tn = 4n + 5.
n
1
2
3
4
5
6
tn
9
13
17
21
25
29
Example 2
Write the first six terms of the sequence defined
by the following formula:
t1 = -2 and tn = 2tn-1 – 1, where n  2
t2 = 2t1 – 1 = 2(-2) – 1 = -5
t3 = 2t2 – 1 = 2(-5) – 1 = -11
t4 = 2t3 – 1 = 2(-11) – 1 = -23
t5 = 2t4 – 1 = 2(-23) – 1 = -47
t6 = 2t5 – 1 = 2(-47) – 1 = -95
Series
A series is an expression that indicates the sum of
terms of a sequence.
Summation notation, which uses the Greek
letter sigma,  , is a way to express a series
in abbreviated form.
5
 2n
n 1
“the sum of 2n for values of n from 1 to 5”
Example 3
Write the terms of each series. Then
evaluate.
6
 2k
k 1
2 + 4 + 6 + 8 + 10 + 12 = 42
Summation Properties
For sequences ak and bk and positive integer n:
n

k =1
n
cak = c  ak
k =1
Example 4
Write the terms of each series. Then
evaluate.
6
 2k
k 1
6
 2 k = 2(1 + 2 + 3 + 4 + 5 + 6)
k 1
= 2(21)
= 42
Practice
Write the first four terms of each sequence.
1) tn = -7n + 3
2) t1 = 0
tn = tn-1 - 4
Summation Properties
For sequences ak and bk and positive integer n:
n

k =1
(ak + bk) =
n

k =1
ak +
n

k =1
bk
Example 1
Write the terms of each series. Then evaluate.
5
5
k 1
k 1
2
2
2k

k

2k

 
5
k
k 1
= (2 + 8 + 18 + 32 + 50) – (1 + 2 + 3 + 4 + 5)
= (110) – (15)
= 95
Summation Formulas
For positive integers n:
Constant Series
n

k =1
c = nc
Example 2
Write the terms of each series. Then evaluate.
4
3
j 1
= (4)(3)
= 12
Summation Formulas
For positive integers n:
Constant Series
Linear Series
n

k =1
n
c = nc
n(n + 1)
 k= 2
k =1
Example 3
Evaluate.
5
 2m
m 1
5
 2 m
m 1
5(5  1)
 (2) 
2
5(6)
 (2) 
2
= -30
Example 3
Evaluate.
25
 2m
m 1
25
 2 m
m 1
25(25  1)
 ( 2) 
2
25(26)
 ( 2) 
2
= -650
Summation Formulas
For positive integers n:
Constant Series
Linear Series
Quadratic Series
n

k =1
c = nc
n
n(n + 1)
 k= 2
k =1
n

k =1
k2
n(n + 1)(2n + 1)
=
6
Example 4
Evaluate.
5
5
m 1
m 1
2
2
12m
 12 m

5(5  1)(2  5  1)
 (12) 
6
(5)(6)(11)
 (12) 
6
= 660
Example 5
Evaluate
5

k =1
(7k2 – 2k + 5) .
5
5
k =1
k =1
= 7  k2 - 2  k +
7
(
5  6  11
6
) – 2(
5

k =1
56
2
5
) + 5  5 = 380
Write the first five terms of each sequence.
1) tn = 6n - 7
2) a1 = -1; an = 3an-1 + 5
Evaluate the sum.
5
2
(2m
 3m  2)
3) 
m 1
Arithmetic Sequences
An arithmetic sequence is a sequence whose
successive terms differ by the same number,
d, called the common difference.
nth Term of an Arithmetic Sequence
tn = t1 + (n – 1)d
tn: nth term
t1: first term
d: common difference
Example 1
Find the 10th term of the sequence defined
by t1 = 7 and tn = tn-1 + 6
tn = t1 + (n – 1)d
t10 = 7 + (10 – 1)6
t10 = 61
Example 2
Find the 15th term of the arithmetic sequence
in which t5 = 7 and t10 = 22.
1. Find the common difference
n
5
6
7
8
9
10
t(n)
7
10
13
16
19
22
d
d
d
7 + 5d = 22
5d = 15
d=3
d
d
Example 2
Find the 15th term of the arithmetic sequence
in which t5 = 7 and t10 = 22.
2. Find t1
tn = t1 + (n – 1)d
7 = t1 + (5 – 1)3
7 = t1 + 12
-5 = t1
Example 2
Find the 15th term of the arithmetic sequence
in which t5 = 7 and t10 = 22.
3. Find the 15th term.
tn = t1 + (n – 1)d
t15 = -5 + (15 – 1)3
t15 = -5 + 42
t15 = 37
Example 3
Find the five arithmetic means between 6 and 60.
15 24 33 42 51
6,___,___,___,___,___,60
tn = t1 + (n – 1)d
60 = 6 + (7 – 1)d
54 = 6d
9=d
Exercises
Find the sum of each series.
1)
4
6
n 1
2)
6
 (k  2)
k 1
3)
5
 4i
i1
4) Find the 12th term of the sequence defined by
t2 = 6 and t8 = 24.
Sum of the First n Terms of an
Arithmetic Series
Sn = n
(
t1 + t n
2
)
Example 1
Given 16,12,8,4,…, find S11.
Sn = n
(
t1 + t n
2
)
(
16-24
2
tn = t1 + (n – 1)d
S11 = 11
t11 = -24
S11 = 11(-4)
t11 = 16 + (11 – 1)(-4)
S11 = -44
)
Example 2
Evaluate
21

k =1
(5 + 4k) .
t1 = 9 and t2 = 13, so d = 4
tn = t1 + (n – 1)d
t21 = 9 + (21 – 1)4
t21 = 89
( )
9 + 89
= 21 (
2 )
Sn = n
S21
t1 + tn
2
S21 = 21(49)
S21 = 1029
Practice
Evaluate
15

k =1
(22 – 7k) .