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Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing this problem.

Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

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Page 1: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Problem:

Add the first 100 counting numbers together.

1 + 2 + …+ 99 + 100

We shall see if we can find a fast way of doing this problem.

Page 2: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

A pattern of numbers in a particular order is called a number sequence, and the individual numbers in the sequence are called terms.

Page 3: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Each number or term in the sequence is associated with a position, which is also a number.

Page 4: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Examples of Sequences

1,2,3,4,5,….

1,1,2,3,5,8,…

1,2,4,8,16,32,…

2,1,7,3,9,3,…

The dots at the end of the sequence indicate that the sequence continues without end. Note not all sequences have a pattern.

Page 5: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

The sequence

2, 5, 8, 11, 14,…

Has first differences of 3 all the time, this makes this sequence arithmetic. If a sequence is arithmetic the first differences must be the same.

Page 6: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

First Differences

First differences are the differences found by subtracting two consecutive numbers in a sequence.

Ex.

2, 5, 8, 11, 14,….

Page 7: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

First Differences

First differences are the differences found by subtracting two consecutive numbers in a sequence.

Ex.

2, 5, 8, 11, 14,….

5 – 2 = 3

3

Page 8: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

First Differences

First differences are the differences found by subtracting two consecutive numbers in a sequence.

Ex.

2, 5, 8, 11, 14,….

8 – 5 = 3

3 3

Page 9: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

First Differences

First differences are the differences found by subtracting two consecutive numbers in a sequence.

Ex.

2, 5, 8, 11, 14,….

11 - 8 = 3

3 3 3

Page 10: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

First Differences

First differences are the differences found by subtracting two consecutive numbers in a sequence.

Ex.

2, 5, 8, 11, 14,….

Since all the first differences are the same this must be an arithmetic sequence

3 3 3 3

Page 11: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Position Numbers

Each term in the sequence is paired with a position number

2, 5, 8, 11, 14,…

Page 12: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

2, 5, 8, 11, 14,…

(recall the first difference is 3)

5 = 2 + 3

Page 13: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

2, 5, 8, 11, 14,…

(recall the first difference is 3)

5 = 2 + 3

8 = 5 + 3 = 2 + 3 + 3

(the underlined part is the previous term)

Page 14: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

2, 5, 8, 11, 14,…

(recall the first difference is 3)

5 = 2 + 3

8 = 5 + 3 = 2 + 3 + 3

11 = 8 + 3 = 2 + 3 + 3 + 3

(the underlined part is the previous term)

Page 15: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

2, 5, 8, 11, 14,…

(recall the first difference is 3)

5 = 2 + 3

8 = 5 + 3 = 2 + 3 + 3

11 = 8 + 3 = 2 + 3 + 3 + 3

14 = 11 + 3 = 2 + 3 + 3 + 3 + 3

(the underlined part is the previous term)

Page 16: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequence: Predictability

The question is how many times does one move from the first term in a sequence.

Ex.

2, 5, 8, 11, 14,….

Take one step from 2 to get to 5

Take two steps from 2 to get to 8

Take three steps from 2 to get to 11

Page 17: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

• Have a constant first difference

• Are predictable

Page 18: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Predictability

If the sequence in question is arithmetic the position numbers will begin with a 0th position.

i.e.,

2, 5, 8, 11, 14, …

Page 19: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Predictability

If the sequence in question is arithmetic the position numbers will begin with a 0th position.

i.e.,

2, 5, 8, 11, 14, …

2 is in the 0th position. Zero 3’s were added to 2 to get to 2. But 2 is the 1st term in the sequence.

Page 20: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Predictability

If the sequence in question is arithmetic the position numbers will begin with a 0th position.

i.e.,

2, 5, 8, 11, 14, …

5 is in the 1st position. One 3 was added to 2 to get to 5. But 5 is the 2nd term in the sequence.

Page 21: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Predictability

If the sequence in question is arithmetic the position numbers will begin with a 0th position.

i.e.,

2, 5, 8, 11, 14, …

8 is in the 2nd position. Two 3’s were added to 2 to get to 8. But 8 is the 3rd term in the sequence.

Page 22: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Predictability

If the sequence in question is arithmetic the position numbers will begin with a 0th position.

i.e.,

2, 5, 8, 11, 14, …

11 is in the 3rd position. Three 3’s were added to 2 to get to 11. But 11is the 4th term in the sequence.

Page 23: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Predictability

If we wanted to know the term in the 22nd position, we would add 22 threes to 2 to get the result, i.e.,

22(3) +2 = 66 + 2 = 68.

68 is in the 22nd position, but it is the 23rd term in the sequence.

Page 24: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Predictability

Mathematics is the study of patterns, and therefore we must look for a pattern.

The position number is the same as the number of first differences that must be added to the first term.

This leads me to believe that the following formula might work.

Term = (Position Number)(Step Size) + Initial Step

Page 25: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

Start with a sequence

-2, 3, 8, 13, 18, 23,…

Find the step size (first difference).

3 – (-2) = 5

Find the initial step (term in 0th position)

-2

Page 26: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

Term = (position)(step size) + initial step

From above

Term= (position)(5) + -2

This is a relationship between the position of a term and the term in a position.

Page 27: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequences

Term = (position)(step size) + initial step

From above

Term= (position)(5) + -2

This is a relationship between the position of a term and the term in a position.

What is the 45th term?

Page 28: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequence

Term = (position)(5) – 2

Term = (45)(5) – 2

Term = 210 – 2

Term = 208

By knowing the position we are able to determine the term, what about the other way around.

Page 29: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequence

What if we know the term but not the position, can the position be determined?

Suppose that 183 is a number in the sequence

–2, 3, 8, 13, 18,…

Page 30: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Arithmetic Sequence

The relationship

Term = (Position)(Step Size) + Initial Step

Replace with known values

183 = (Position)(5) – 2

37 = Position

Page 31: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

Find the sum of the sequence:

-2, 3, 8, 13, …,603, 608, 613

Page 32: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Is the same as

S = 613 + 608 + 603 + … + 8 + 3 + (-2)

Page 33: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)

Add down in columns defined by the plus signs.

Page 34: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)2S = 611 + 611+ 611 + … + 611 + 611 + 611

If we knew the number of terms in the original sequence we could answer the question 611+ 611 + … + 611 + 611.

Page 35: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)2S = 611 + 611+ 611 + … + 611 + 611 + 611

The sequence –2, 3, 8, 13,…

Is arithmetic therefore we have predictability.

Page 36: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)2S = 611 + 611+ 611 + … + 611 + 611 + 611

613 = (Position)(5) – 2

123 = Position, which means there are 124 terms in the sequence.

Page 37: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)2S = 611 + 611+ 611 + … + 611 + 611 + 611

There are 124 - 611’s in the sequence

Page 38: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)2S = 611 + 611+ 611 + … + 611 + 611 + 6112S = (611)(124)

Page 39: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)2S = 611 + 611+ 611 + … + 611 + 611 + 6112S = (611)(124)2S = 75764

Page 40: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

S = -2 + 3 + 8 + … + 603 + 608 + 613

Notice the following

S = 613 + 608 + 603 + … + 8 + 3 + (-2)2S = 611 + 611+ 611 + … + 611 + 611 + 6112S = (611)(124)2S = 75764S = 37882

Page 41: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

Let us find the sum of the sequence

5, 9, 13, … , 365, 369, 373

First we must find out how many terms are in the sequence. In order to do this we must determine the step size and initial step of the sequence.

Page 42: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

Let us find the sum of the sequence

5, 9, 13, … , 365, 369, 373

Step Size = 4

Initial Step = 5

Hence

Term = 4(Position) + 5

Page 43: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

New Problem:

Let us find the sum of the sequence

5, 9, 13, … , 365, 369, 373

Term = 4(Position) + 5

373 = 4(Position) + 5

92 = Position

We will need this information later.

Page 44: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Step 1: Write down the sum you wish to determine

S = 5 + 9 + 13 + … + 365 + 369 + 373

Page 45: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Step 2: reverse the order of the sum and write it under the first sum.

S = 5 + 9 + 13 + … + 365 + 369 + 373S = 373 + 369 + 365 + …+ 13 + 9 + 5

Page 46: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Step 3: add down in columns determined by the plus signs in the problem.

S = 5 + 9 + 13 + … + 365 + 369 + 373S = 373 + 369 + 365 + …+ 13 + 9 + 5

2S = 378 + 378 + 378+ … + 378 + 378 + 378

Notice that all the columns have the same sum, if only we knew how many they were!

Page 47: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Step 4: recall we had predictability with arithmetic sequences and we already determined that 373 was in position 92 of the

original sequence. Therefore 93 terms in the sequence.

S = 5 + 9 + 13 + … + 365 + 369 + 373S = 373 + 369 + 365 + …+ 13 + 9 + 5

2S = 378 + 378 + 378+ …+ 378 + 378 + 378

Page 48: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Step 5: replace the long sum with the associated multiplication problem.

S = 5 + 9 + 13 + … + 365 + 369 + 373S = 373 + 369 + 365 + …+ 13 + 9 + 5

2S = 378 + 378 + 378+ …+ 378 + 378 + 3782S = 378(93)

Page 49: Designed by David Jay Hebert, PhD Problem: Add the first 100 counting numbers together. 1 + 2 + …+ 99 + 100 We shall see if we can find a fast way of doing

Step 6: solve for S and be done with the problem.

S = 5 + 9 + 13 + … + 365 + 369 + 373S = 373 + 369 + 365 + …+ 13 + 9 + 5

2S = 378 + 378 + 378+ …+ 378 + 378 + 3782S = 378(93)2S = 35154

S = 17577