TASK 4:Sequences,

Mathematical Induction and Recursion

Definition :

ordered set of mathematical objects.

SEQUENCES

Find the sum of all the integers from 1 to 1000.

The sequence of integers starting from 1 to 1000 is given by 

1 , 2 , 3 , 4 , ... , 1000 

The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above). 

s1000 = 1000 (1 + 1000) / 2 = 500500

EXAMPLE PROBLEM

Sequeance

Arithmetic Sequences

Geometric Sequences

Special Integers

Sequence

TYPE OF SEQUENCE

In an Arithmetic Sequence the difference between one term and the next is a constant.

In other words, you just add the same value each time ... infinitely.

WHAT IS ARITHMETIC SEQUENCE

In General you could write an arithmetic sequence like this:

{a, a+d, a+2d, a+3d, ... }

where:

a is the first term, and

d is the difference between the terms (called the "common difference")

In a Geometric Sequence each term is found by multiplying the previous term by a constant.

In General you could write a Geometric Sequence like this:

{a, ar, ar2, ar3, ... }

where:

a is the first term, and

r is the factor between the terms (called the "common ratio")

Geometric Sequence

• This Triangular Number Sequence is generated from a pattern of dots which form a triangle.

Triangular Numbers

• Square numbers, better known as perfect squares, are an integer which is the product of that integer with itself.

Square Numbers

• The Fibonacci Sequence is found by adding the two numbers before it together. 

Fibonacci Numbers

• A cube number sequence is a mathematical sequence consisting of a sequence in which the next term originates by multiplying the number 3 times with itself, or in other words, raising it to the power of three

Cube Numbers

SPECIAL INTEGER SEQUENCE

An example of this type of number sequence could be the following:

1,   3,   6,   10,   15,   21,   28,   36,   45,   …

This sequence is generated from a pattern of dots which form a triangle. By adding another row of dots and counting all the dots we can find the next number of the sequence.

Triangular Numbers

Continue….

1,   4,   9,   16,   25,   36,   49,   64,   81,   …

The next number is made by squaring where it is in the pattern. 

The second number is 2 squared (22 or 2×2) 

The seventh number is 7 squared (72 or 7×7) etc

Square Numbers

0,   1,   1,   2,   3,   5,   8,   13,   21,   34,   …

The Fibonacci Sequence is found by adding the two numbers before it together. 

The 2 is found by adding the two numbers before it (1+1) 

The 21 is found by adding the two numbers before it (8+13) 

The next number in the sequence above would be 55 (21+34)

Fibonacci Numbers

1,   8,   27,   64,   125,   216,   343,   512,   729,   …

The next number is made by cubing where it is in the pattern. 

The second number is 2 cubed (23 or 2×2×2) 

The seventh number is 7 cubed (73 or 7×7×7) etc

Cube Numbers

Give 1 example sequences use in computer programming

Sequence in computer programming is the default control structure, instructions are executed one after another. They might, for example, carry out a series of arithmetic operations, assigning results to variables

 

Example:-

Make a list number of 100 students that attend “seminar keusahawanan”.

 

Pseudocode

BEGIN

Declare a list as an array with 100 elements

for (int student = 0, student<100; student++){

list[student] = student + 1 (arithmetic operation)

print list[student]

}

END

Answer:

No.Student = 0

Print list[student] = 1

No.Student = 1

Print list[student] = 2

.

.

.

.

Print list[student] = 99

No.Student = 100

Source Code

public class sequences {

public static void main(String[] args) {

int list [] = new int[100];

for(int student = 0; student<100; student++) {

list[student] = student + 1;

System.out.println(list[student]);

}

}

 

}

PRINCIPLE OF MATHEMATICAL INDUCTION

To prove that P(n) is true for all positive integers n, where P(n) is a propositional funtion

METHOD OF PROOF

1) BASIS STEP: We verify that P(1) is true.

2) INDUCTIVE STEP: We show that the conditional statement

P(k) → P(k + 1) is true for all positive

integers k.

PRINCIPLE OF MATHEMATICAL INDUCTION &

METHOD OF PROOF

Problem1:

Show that if n is a positive integer, then

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

BASIC STEP: P(1),

1 = 1(1 + 1)/2

1 =1 (RHS=LHS)

INDUCTIVE STEP: we assume that 1 + 2 + ... + k = k(k + 1)/2

it must be shown that P(k + 1) is true,

1 + 2 + ... + k + ( k +1) = (k + 1)[( k+ 1) (k +1) + 1] /2 = (k + 1)(k + 2)/ 2

Add k +1 to both sides of the equation P(k)

1 + 2 + ... + k + (k +1) = k(k +1) /2 + (k +1)

= k(k +1 ) + 2 (k +1 )/2

= (k + 1) (k + 2)/2

Problem 2:

For all n≥0

= n(n+1)(2n+1)/6

Basis:

=0

Induction:

= [n+1][(n+1)+1][2(n+1)+1]/6

= (2n3 +3n +n)/6

= (2n3 +9n +13n+6)/6

() + (n+1) = (2n3 +9n +13n+6)/6

(2n3 +3n +n)/6 + n +2n+1 = (2n3 +9n +13n+6)/6