designed by david jay hebert, phd more on sequences some sequences are not arithmetic and therefore...

More on Sequences

Some sequences are not arithmetic and therefore begin with a 1st position.

Some sequences come from patterns that occur in nature, and some from patterns of figures.

These are the triangular figures and give rise to triangular numbers.

1, 3, 6, 10, 15, …

1, 3, 6, 10, 15, …

What is the pattern?

1, 3, 6, 10, 15, …

What is the pattern?

Look at the first differences.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

The first differences are not all the same, but the set of first differences should look familiar. Find the second differences.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

1, 1, 1, …

The second differences are all the same; therefore, the first differences are arithmetic.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Notice the following relationships:

1 = 1

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Notice the following relationships:

1 = 1

3 = 1 + 2

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Notice the following relationships:

1 = 1

3 = 1 + 2

6 = 1 + 2 + 3

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Notice the following relationships:

1 = 1

3 = 1 + 2

6 = 1 + 2 + 3

10 = 1 + 2 + 3 + 4

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Notice the following relationships:

1 = 1

3 = 1 + 2

6 = 1 + 2 + 3

10 = 1 + 2 + 3 + 4

A triangular number is nothing more that the sum of an arithmetic sequence.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Recall that the sequence is not arithmetic and therefore begins with position 1.

1, 3, 6, 10, 15, …

1st 2nd 3rd 4th 5th

These are the position numbers.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

The 1st triangular number is 1.

The 2nd triangular number is 1 + 2.

The 3rd triangular number is 1 + 2 + 3, etc…

1, 3, 6, 10, 15, …2, 3, 4, 5, …

What is the 17th triangular number?

1, 3, 6, 10, 15, …2, 3, 4, 5, …

What is the 17th triangular number?

T17 = 1 + 2 + 3 + … + 15 + 16 + 17

1, 3, 6, 10, 15, …2, 3, 4, 5, …

What is the 17th triangular number?

T17 = 1 + 2 + 3 + … + 15 + 16 + 17

T17 = 17 + 16 + 15 + …+ 3 + 2 + 1

1, 3, 6, 10, 15, …2, 3, 4, 5, …

What is the 17th triangular number?

T17 = 1 + 2 + 3 + … + 15 + 16 + 17

T17 = 17 + 16 + 15 + …+ 3 + 2 + 12T17 = 18 + 18 + 18 + …+18 + 18 + 18

1, 3, 6, 10, 15, …2, 3, 4, 5, …

What is the 17th triangular number?

T17 = 1 + 2 + 3 + … + 15 + 16 + 17

T17 = 17 + 16 + 15 + …+ 3 + 2 + 12T17 = 18 + 18 + 18 + …+18 + 18 + 182T17 = 18(17)

1, 3, 6, 10, 15, …2, 3, 4, 5, …

What is the 17th triangular number?

T17 = 1 + 2 + 3 + … + 15 + 16 + 17

T17 = 17 + 16 + 15 + …+ 3 + 2 + 12T17 = 18 + 18 + 18 + …+18 + 18 + 182T17 = 18(17)T17 = 153

1, 3, 6, 10, 15, …2, 3, 4, 5, …

A more general question is what does a general triangular number look like.

We must be able to describe any number we want in an arithmetic sequence.

If an arithmetic sequence has a step size of 4 and an initial step of 5, then the number in the 12th position (the 13th term in the sequence) will be

4(12) + 5

We must be able to describe any number we want in an arithmetic sequence.

If an arithmetic sequence has a step size of 4 and an initial step of 5, then the number in the 12th position (the 13th term in the sequence) will be

4(12) + 5

If an arithmetic sequence has a step size of 4 and an initial step of 5, then the number in the 23rd position (the 24th term in the sequence) will be

4(23) + 5

We must be able to describe any number we want in an arithmetic sequence.

What would be the nth term in the sequence?

4n + 5

This means there are n + 1 terms in the entire sequence to this point.

What would be the term just before the nth term?

4(n – 1) + 5 = 4n + 1

1, 3, 6, 10, 15, …2, 3, 4, 5, …

A more general question is what does a general triangular number look like.

In order to answer this question we must first find the general term of the sequence of first differences.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Since the first differences are 1 and the initial step is 1*, then the general nth term is 1(n-1) + 1 or n.

We begin at 1 since the sums we want begin with a one, i.e., 10 = 1 + 2 + 3 + 4

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Now let us answer the question what is

1 + 2 + 3 + … + + + ( n ).

Let us fill in the blanks.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

Now let us answer the question what is

1 + 2 + 3 + … + (n-2) + (n-1) + ( n ).

How to answer the above question.

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = 1 + 2 + 3 + … + (n-2) + (n-1) + ( n )

S = ( n ) + (n-1) + (n-2) + … + 3 + 2 + 1

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = 1 + 2 + 3 + … + (n-2) + (n-1) + ( n )

S = ( n ) + (n-1) + (n-2) + … + 3 + 2 + 12S = (n+1) + (n+1) + (n+1) + … + (n+1)+ (n+1) + (n+1)

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = 1 + 2 + 3 + … + (n-2) + (n-1) + ( n )

S = ( n ) + (n-1) + (n-2) + … + 3 + 2 + 12S = (n+1) + (n+1) + (n+1) + … + (n+1)+ (n+1) + (n+1)2S = n(n + 1)

S = n(n + 1)2

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = n(n + 1)2

What does this formula do you ask. Watch this!

n =1 S = 1(1 + 1)/ 2 = 2/2 = 1

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = n(n + 1)2

What does this formula do you ask. Watch this!

n =1 S = 1(1 + 1)/ 2 = 2/2 = 1

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

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = n(n + 1)2

What does this formula do you ask. Watch this!

n =1 S = 1(1 + 1)/ 2 = 2/2 = 1

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

n = 3 S = 3(3 + 1)/2 = 3(4)/2 = 6

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = n(n + 1)2

What does this formula do you ask. Watch this!

n =1 S = 1(1 + 1)/ 2 = 2/2 = 1

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

n = 3 S = 3(3 + 1)/2 = 3(4)/2 = 6

n = 4 S = 4(4 + 1)/2 = 4(5)/2 = 10

1, 3, 6, 10, 15, …2, 3, 4, 5, …

S = n(n + 1)2

This formula generates triangular numbers starting with a 1st position.

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

2, 2, ….. (second differences)

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

Notice that

1 = 1 1st square number (0th term sum)

4 = 1 + 3 2nd square number (1st term sum)

9 = 1 + 3 + 5 3rd square number (2nd term sum)

16 = 1 + 3 + 5 + 7, etc….

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

To solve the problem look at the set of first differences, and find a sum of the first n terms.

In order to do this we must first describe the term in the nth position of the sequence of first differences.

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

To solve the problem look at the set of first differences, and find a sum of the first n terms.

In order to do this we must first describe the term in the nth position of the sequence of first differences.

2n + 1

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

To solve the problem look at the set of first differences, and find a sum of the first n terms or the term in position n - 1. Because arithmetic sequences from a 0th position

2(n – 1) + 1 = 2n – 2 + 1 = 2n - 1

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

So any square number is described by the sum:

S = 1 + 3 + … +(2n – 5) + (2n – 3) + (2n – 1)

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

How to sum such a sequence, Gauss’s trick

S = 1 + 3 +… + (2n – 3) + (2n – 1)

S = (2n – 1) + (2n – 3) + … + 3 + 1

1, 4, 9, 16,…. (square numbers)3, 5, 7, ….. (first differences)

How to sum such a sequence, Gauss’s trick

S = 1 + 3 +… + (2n – 3) + (2n – 1)

S = (2n – 1) + (2n – 3) + … + 3 + 12S = (2n) + (2n) + …. + (2n) + (2n)

2S = (2n)(n)

2S = 2n2

S = n2

1, 5, 12, 22,…

Are the pentagonal numbers. What is the pattern?

1, 5, 12, 22,…

First differences: 4 , 7, 10,…

Second differences: 3, 3, …..

1, 5, 12, 22,…

First differences: 4 , 7, 10,…

Notice that

1 = 1

5 = 1 + 4

12 = 1 + 4 + 7, etc…

1, 5, 12, 22,…

First differences: 4 , 7, 10,…

General term for the first differences is

3n + 1.

If we want to add n numbers of this form add from the 0th to the (n –1)th terms.

1, 5, 12, 22,…

P = 1 + 4 + … + (3n – 5) + (3n – 2)

P = (3n – 2) + (3n – 5) + … + 4 + 1

1, 5, 12, 22,…

P = 1 + 4 + … + (3n – 5) + (3n – 2)

P = (3n – 2) + (3n – 5) + … + 4 + 1

2P = (3n – 1) + (3n – 1) + … +(3n – 1) + (3n – 1)

2P = (3n – 1)n

P = (3n – 1)n2