Examples: Principle of Mathematical Induction - 2

# Examples: Principle of Mathematical Induction - 2 Video Lecture | Mathematics (Maths) Class 11 - Commerce

## Mathematics (Maths) Class 11

85 videos|243 docs|99 tests

## FAQs on Examples: Principle of Mathematical Induction - 2 Video Lecture - Mathematics (Maths) Class 11 - Commerce

 1. What is the Principle of Mathematical Induction?
Ans. The Principle of Mathematical Induction is a proof technique used to establish that a given statement or property is true for all natural numbers. It consists of two steps: the base step and the inductive step. In the base step, we show that the statement holds true for the first natural number, usually 1. In the inductive step, we assume that the statement is true for some arbitrary natural number, and then prove that it holds true for the next natural number. By repeating the inductive step, we can conclude that the statement holds true for all natural numbers.
 2. How is the Principle of Mathematical Induction used in JEE exams?
Ans. The Principle of Mathematical Induction is frequently tested in JEE exams, particularly in the mathematics section. Questions related to sequences, series, divisibility, inequalities, and algebraic expressions often require the application of mathematical induction. It is important to understand the concept and practice solving problems using this technique to perform well in the JEE exams.
 3. Can you explain an example of using the Principle of Mathematical Induction?
Ans. Sure! Let's consider the statement: "For every natural number n, 1 + 2 + 3 + ... + n = n(n+1)/2." To prove this using mathematical induction, we start with the base step: When n = 1, the left-hand side (LHS) is 1, and the right-hand side (RHS) is 1(1+1)/2 = 1. Hence, the statement holds true for n = 1. In the inductive step, we assume that the statement is true for some arbitrary natural number k, i.e., 1 + 2 + 3 + ... + k = k(k+1)/2. Now, we need to prove that the statement holds true for k+1, i.e., 1 + 2 + 3 + ... + k + (k+1) = (k+1)((k+1)+1)/2. Using the assumption, we can rewrite the LHS as k(k+1)/2 + (k+1). By simplifying, we get (k+1)((k/2) + 1) = (k+1)(k+2)/2, which is equal to the RHS. Therefore, the statement holds true for k+1. By the principle of mathematical induction, we can conclude that the statement is true for all natural numbers.
 4. Are there any common mistakes to avoid when using the Principle of Mathematical Induction?
Ans. Yes, there are a few common mistakes to avoid when using the Principle of Mathematical Induction. One common mistake is assuming that the statement is true for the next natural number without properly proving it. It is important to clearly demonstrate the steps and logic used to reach the conclusion for each natural number. Another mistake is not specifying the base step properly or assuming that the statement is true for all natural numbers without verifying it for the initial value. It is crucial to establish the truth of the statement for the first natural number before proceeding with the inductive step.
 5. Can the Principle of Mathematical Induction be used for proving statements in other number systems?
Ans. No, the Principle of Mathematical Induction is specific to the natural numbers. It cannot be directly applied to prove statements in other number systems such as integers, rational numbers, or real numbers. However, variations of mathematical induction, such as strong induction, can be used to prove statements in other number systems. Strong induction allows assuming that the statement is true for all natural numbers up to a certain value, rather than just the previous natural number. This extension of mathematical induction can be used to prove statements in other number systems.

## Mathematics (Maths) Class 11

85 videos|243 docs|99 tests

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