1st Year Math Notes Chapter 8
Important Complete 1st Year Math Notes Chapter 8 of Mathematical Induction and Binomial Theorem written by Professor M. Sulman Sherazi Suib. These notes are very helpful in the preparation of Mathematical Induction and Binomial Theorem class 11 for the students of Mathematics of the Intermediate and these are according to the paper patterns of all Punjab boards.
Summary and Contents:
Topics which are discussed in the notes are given below:
- Our comprehensive 1st Year Math Notes Chapter 8 will ensure you're fully prepared for your exams.
- Mathematical induction and principle of Mathematical Induction with examples.
- Principle of extended Mathematical Induction with examples.
- State and prove the binomial theorem
- What is a binomial expression with its calculation
- The Binomial Theorem when the index n is a negative integer or a fraction.
- Francesco Mourolico (1494-1575) devised the method of induction and applied this device irst to prove that the sum of the irst n odd positive integers equals n2. He presented many properties of integers and proved some of these properties using the method of mathematical induction. We are aware of the fact that even one exception or case to a mathematical formula is enough to prove it to be false. Such a case or exception which fails the mathematical formula or statement is called a counter example. The validity of a formula or statement depending on a variable belonging to a certain set is established if it is true for each element of the set under consideration. For example, we consider the statement S(n) = n2 - n + 41 is a prime number for every natural number n. The values of the expression n2 - n + 41 for some irst natural numbers are given in the table as shown below:
- Principle of Mathematical Induction: The principle of mathematical induction is stated as follows: If a proposition or statement S(n) for each positive integer n is such that 1) S(1) is true i.e., S(n) is true for n = 1 and 2) S(k + 1) is true whenever S(k) is true for any positive integer k, then S(n) is true for all positive integers.
- Procedure: Substituting n = 1, show that the statement is true for n = 1. Assuming that the statement is true for any positive integer k, then show that it is true for the next higher integer. For the second condition, one of the following two methods can be used: M1 Starting with one side of S(k +1), its other side is derived by using S(k). M2 S(k + 1) is established by performing algebraic operations on S(k).
- Care should be taken while applying this method. Both the conditions (1) and (2) of the principle of mathematical induction are essential. The condition (1) gives us a starting point but the condition (2) enables us to proceed from one positive integer to the next. In the condition (2) we do not prove that S(k + 1) is true but prove only that if S(k)is true, then S(k + 1) is true. We can say that any proposition or statement for which only one condition is satisied, will not be truefor all n belonging to the set of positive integers. For example, we consider the statement that 3n is an even integer for any positive integer n. Let S(n) be the given statement. Assume that S(k) is true, that is, 3k in an even integer for n = k. When 3k is even, then 3k + 3k + 3k is even which implies that 3k.3= 3k +1 is even. This shows that S(k + 1 ) will be true when S(k) is true. But 31 is not an even integer which relects that the irst condition does not hold. Thus our supposition is false.
- Sometimes, we wish to prove formuIae or statements which are true for all integers n greater than or equal to some integer i, where i m1. In such cases, S(1) is replaced by S(i) and the condition (2) remains the same. To tackle such situations, we use the principle of extended mathematical induction which is stated as below.