Why is the base case $P(1)$ non-negotiable in Mathematical Induction?

Without $P(1)$, the chain reaction never begins. For example, the statement $n = n + 1$ satisfies $P(k) \implies P(k+1)$ (since $k = k+1 \implies k+1 = k+2$), but it is false for every $n \in \mathbb{N}$ because there is no base case $P(1)$.

Can the Principle of Mathematical Induction be applied to statements starting at $n = 4$?

Yes, this is Extended Mathematical Induction. The base case starts at $P(n_0)$ (here $P(4)$), proving the statement true for all integers $n \ge n_0$.

What is the quickest strategy to solve PMI questions in objective JEE Main MCQs?

For objective MCQs in JEE Main, test small values such as $n = 1$, $n = 2$, and $n = 3$. Eliminate options that fail for these values. However, for structural questions asking about the transition from $P(k)$ to $P(k+1)$, formal algebraic expansion is required.

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Solved High-Yield Sample Questions

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Question #1JEE Main 2024

For all natural numbers $n \in \mathbb{N}$ , the expression $n^3 + 2n$ is always divisible by which of the following integers?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Check the base case for $n = 1$ . $P(1) = 1^3 + 2(1) = 3$ , which is divisible by $3$ .

Step 2

Assume $P(k)$ is true for $n = k$ , i.e., $k^3 + 2k = 3\lambda$ for some integer $\lambda \in \mathbb{Z}$ .

Step 3

Evaluate $P(k+1) = (k+1)^3 + 2(k+1) = k^3 + 3k^2 + 3k + 1 + 2k + 2 = (k^3 + 2k) + 3(k^2 + k + 1) = 3\lambda + 3(k^2 + k + 1) = 3(\lambda + k^2 + k + 1)$ , which is clearly divisible by $3$ .

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Question #2JEE Main 2023

If $P(n)$ is the statement ' $2^{2n} - 1$ is divisible by $3$ ', then in proving $P(k) \implies P(k+1)$ , the expression $2^{2(k+1)} - 1$ is written in terms of $2^{2k} - 1$ as:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Expand $2^{2(k+1)} - 1$ algebraically to get $2^{2k+2} - 1 = 4 \cdot 2^{2k} - 1$ .

Step 2

Add and subtract $4$ to isolate the inductive hypothesis term $(2^{2k} - 1)$ , yielding $4 \cdot 2^{2k} - 4 + 3$ .

Step 3

Factor out $4$ to get $4(2^{2k} - 1) + 3$ . Since $2^{2k}-1$ is divisible by $3$ , $4(2^{2k}-1) + 3$ is also divisible by $3$ .

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Mastering Principle of Mathematical Induction: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Base Step & Inductive Step Paradigm

Sum of First $n$ Natural Numbers

Divisibility Formulation

Bernoulli's Inequality

Sum of Squares of First $n$ Natural Numbers

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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