Designed by David Jay Hebert, PhD

Name: Designed by David Jay Hebert, PhD
Uploaded: 2018-01-09T14:51:22+00:00
Duration: PTM24S59
Channel: Michael Cole
Description: Designed by David Jay Hebert, PhD

Designed by David Jay Hebert, PhD
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. Designed by David Jay Hebert, PhD

These are the triangular figures and give rise to triangular numbers. Designed by David Jay Hebert, PhD

1, , , , , … Designed by David Jay Hebert, PhD

1, , , , , … What is the pattern? Designed by David Jay Hebert, PhD

1, , , , , … What is the pattern? Look at the first differences. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … The first differences are not all the same, but the set of first differences should look familiar. Find the second differences. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … 1, , , … The second differences are all the same; therefore, the first differences are arithmetic. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Notice the following relationships: 1 = 1 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Notice the following relationships: 1 = 1 3 = 1 + 2 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Notice the following relationships: 1 = 1 3 = 1 + 2 6 = Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Notice the following relationships: 1 = 1 3 = 1 + 2 6 = 10 = Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Notice the following relationships: 1 = 1 3 = 1 + 2 6 = 10 = A triangular number is nothing more that the sum of an arithmetic sequence. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Recall that the sequence is not arithmetic and therefore begins with position 1. Designed by David Jay Hebert, PhD

1, , , , , … 1st nd rd th th These are the position numbers. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … The 1st triangular number is 1. The 2nd triangular number is The 3rd triangular number is , etc… Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What is the 17th triangular number? Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What is the 17th triangular number? T17 = … Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What is the 17th triangular number? T17 = … T17 = … Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What is the 17th triangular number? T17 = … T17 = … 2T17 = … Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What is the 17th triangular number? T17 = … T17 = … 2T17 = … 2T17 = 18(17) Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What is the 17th triangular number? T17 = … T17 = … 2T17 = … 2T17 = 18(17) T17 = 153 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … A more general question is what does a general triangular number look like. Designed by David Jay Hebert, PhD

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 Designed by David Jay Hebert, PhD

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 Designed by David Jay Hebert, PhD

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 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … 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. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … 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 = Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Now let us answer the question what is … ( n ). Let us fill in the blanks. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … Now let us answer the question what is … + (n-2) + (n-1) + ( n ). How to answer the above question. Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … S = … + (n-2) + (n-1) + ( n ) S = ( n ) + (n-1) + (n-2) + … Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … S = … + (n-2) + (n-1) + ( n ) S = ( n ) + (n-1) + (n-2) + … 2S = (n+1) + (n+1) + (n+1) + … + (n+1)+ (n+1) + (n+1) Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … S = … + (n-2) + (n-1) + ( n ) S = ( n ) + (n-1) + (n-2) + … 2S = (n+1) + (n+1) + (n+1) + … + (n+1)+ (n+1) + (n+1) 2S = n(n + 1) n(n + 1) S = 2 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What does this formula do you ask. Watch this! n(n + 1) S = 2 n =1 S = 1(1 + 1)/ 2 = 2/2 = 1 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What does this formula do you ask. Watch this! n(n + 1) S = 2 n =1 S = 1(1 + 1)/ 2 = 2/2 = 1 n = 2 S = 2(2 + 1)/2 = 2(3)/2 = 3 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What does this formula do you ask. Watch this! n(n + 1) S = 2 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 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … What does this formula do you ask. Watch this! n(n + 1) S = 2 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 Designed by David Jay Hebert, PhD

1, , , , , … 2, , , , … This formula generates triangular numbers starting with a 1st position. n(n + 1) S = 2 Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (first differences) 2, , … (second differences) Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (first differences) Notice that = st square number (0th term sum) = nd square number (1st term sum) = rd square number (2nd term sum) 16 = , etc…. Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (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. Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (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 Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (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 Because arithmetic sequences from a 0th position 2(n – 1) + 1 = 2n – = 2n - 1 Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (first differences) So any square number is described by the sum: S = … +(2n – 5) + (2n – 3) + (2n – 1) Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (first differences) How to sum such a sequence, Gauss’s trick S = … + (2n – 3) + (2n – 1) S = (2n – 1) + (2n – 3) + … Designed by David Jay Hebert, PhD

1, , , ,…. (square numbers) 3, , , … (first differences) How to sum such a sequence, Gauss’s trick S = … + (2n – 3) + (2n – 1) S = (2n – 1) + (2n – 3) + … 2S = (2n) (2n) + …. + (2n) (2n) 2S = (2n)(n) 2S = 2n2 S = n2 Designed by David Jay Hebert, PhD

1, , , ,… Are the pentagonal numbers. What is the pattern? Designed by David Jay Hebert, PhD

1, , , ,… First differences: 4 , , ,… Second differences: , , ….. Designed by David Jay Hebert, PhD

1, , , ,… First differences: 4 , , ,… Notice that 1 = 1 5 = 1 + 4 12 = , etc… Designed by David Jay Hebert, PhD

1, , , ,… First differences: 4 , , ,… 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. Designed by David Jay Hebert, PhD

1, , , ,… P = … + (3n – 5) + (3n – 2) P = (3n – 2) + (3n – 5) + … Designed by David Jay Hebert, PhD

1, , , ,… P = … + (3n – 5) + (3n – 2) P = (3n – 2) + (3n – 5) + … 2P = (3n – 1) + (3n – 1) + … +(3n – 1) + (3n – 1) 2P = (3n – 1)n (3n – 1)n P = 2 Designed by David Jay Hebert, PhD

Designed by David Jay Hebert, PhD

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