Series, Sigma and Arithmetic Sequences and Series

I.  Return Quiz 3

II.  Homework

III. Definition of a Series

Example:  Consider the sequence

a_n = n²  

{1,4,9,16,25,...} 

We make the following definition:  

S₁ = a₁ = 1

S₂ = a₁ + a₂ = 1 + 5 = 5

S₃ =   a₁ + a₂ + a₃ = 1 + 4 + 9 = 13

Have the students find S₄  and S₅ 

In general., for a sequence {a_n}  we define a new sequence

S_n =  a₁ + a₂ + a₃ + ... + a_n  

The class will find S₅ for

A) a_n = 2n + 1 and

B) a_n = (-1)ⁿ     

IV.  Sigma Notation

Instead of using the ... notation we use the following notation:

Example:

We will do the example:

We read this as "The sum from i equals 3 to 6 of 1 over i."  i is called the index of summation.  Think of sigma as a big plus sign.  The bottom number tells you where to start and the top number tells you where to end.

We will do:

To go form a listed sum to sigma notation we follow the steps below:

Example:  3 + 5 + 7 + 9 + ... + 23

Step 1)  Identify a_n  (this goes to the right of the sigma sign)

We see that a_n = 2n + 1

Step 2)  Solve for n to find out what n the first term uses.  (this goes on the bottom of the sigma sign)

Since 2n+ 1 = 3  has solution n = 1, the first term uses n = 1

Step 3)  Find out what n the last term uses (this goes on the top of the sigma sign)

We solve: 2n + 1 = 23 has solution n = 12

Step 4)  Write

The class will try:

{3,6,9,12,...120}and

{-1,2,-4,...,128}

V.  Arithmetic Sequences

Exercise:  Find the next term and the general formula for the following:

{2,5,8,11,14,...}

{0,4,8,12,16,...}

{2,-1,-4,-7,-10,...}

For each of these three sequences there is a common difference.  In the first sequence the common difference is d = 3, in the second sequence the common difference is d = 4, and on the third sequence the common difference is d = -3.  We will call a sequence an arithmetic sequence if there is a common difference.  

The general formula for an arithmetic sequence is

a_n = a₁  + (n - 1)d

What is the difference between the fourth and the tenth terms of

{2,6,10,14,...)

We have

a₁₀ - a₄  = (10 - 4)d = 6(4) = 24.

III.  Arithmetic Series

First we see that

1+ 2 + 3 + ... + 100 = 101 + 101 + ... + 101 (50 times) = 101(50)

In general

1 + 2 + 3 + ... + n = n(n + 1)/2

Example:  What is

S = 1 + 4 + 7 + 10 + 13 +... +  46

Solution:  

S = 1 + (1 + 1(3)) + (1 + 2(3))  + (1 + 3(3)) + ... + (1 + 15(3))

= (1 + 1 + ... + 1) + 3(1 + 2 + 3 + ... + 15)

= 16 + 3(15)(16)/2

In General

S_n = n (a₁)+ d(n - 1)(n)/2 = 1/2[2n(a₁) + d(n - 1)(n)] = 1/2[2n (a₁)+ dn² - dn]

= (n/2)[2 (a₁)+ dn - d]= (n/2)[2 (a₁) + d(n - 1)]

Or Alternatively

S = n/2(a₁ + a_n)

Example:  How much will I receive over my 35 year career if my starting salary is $40,000, and I receive a 1,000 salary raise for each year I work here?

Solution:  We have the series:

40,000 + 41,000 + 42,000 + ... + 74,000

= 35/2 (40,000 + 74,000) = $1,995,500