JEE Main Study Notes on Mathematical Induction: Mathematical Induction is a technique that helps to prove a formula, a statement, or a theorem is true for every natural number. It proves that the statement is true for the iteration number n, then for n+1 number of iteration, the statement is true.

• Candidates can expect around 1-2 questions in JEE Main on the topic of Mathematical Induction.
• JEE Main Mathematics Question Paper has questions about the applications of Mathematical Induction.

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Principle of Mathematical Induction

Mathematical Induction uses 2 steps to prove a given statement whether it is true or not.

First Principle of Mathematical Induction

• Base case: For the first natural number the given statement is true. i.e., for n=1, p(1) is true.

• Inductive Step: If the given statement is true for any natural number like n = k then it will be correct for n = k + 1 also that is if p (k) is true then p (k+1) will also be true.

Examples: Prove that for any natural number n, the sum of n natural numbers is

Solution: Given the statement which we need to prove

Read: JEE Main Probability Study Notes

Step 1: Base Step

Show the statement holds for n =1

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

As the LHS = RHS

This shows that p (1) is true.

Step 2: Inductive Step

Show if p (k) is right then p (k+1) is also right.

Assume p (k) is right

Let n = k

Now we have to prove that it is true for the next number also i.e. k+1

Let n = k+1

Using the induction hypothesis which assumes that p (k) is true, we can rewrite it as:

The principle of mathematical induction shows that if both the above steps are true then p (n) is true for all n natural numbers.

This shows that p (k+1) is true.

The principle of mathematical induction shows that if both the above steps are true then p (n) is true for all n natural numbers.

Example 2

Prove using mathematical induction that

12 + 22 + 32 + · · · + n2 = n (n + 1) (2n + 1)/6 ,for all positive integers n.

Solution: Now we will use base step and induction step to prove it.

Step 1

Let n = 1

The formula is true for n = 1 that is, the statement is true for p (1)

Step 2

Assume p (k) and prove for p (k+1).

Use P(k) to show that P(k + 1) is true.

Therefore, by the Principle of Mathematical Induction,

12 + 22 + 32 + · · · + n2 = n (n + 1) (2n + 1) / 6, is true for all positive integers n.

Example 3

Solution:

Step 1

Show p (1) is true.

So p (1) is true.

Step 2

Assume p (k) is true.

Show if p (k) is true then p (k+1) is also true.

Now use p (k) and p (1) to prove it.

This shows that this statement is true for p (k+1) also.

Hence p (n) is true.

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Second Principle of Mathematical Induction

The second principle of Mathematical Induction is called strong induction because this induction is more powerful than the first principle. It doesn’t need a base case in a special case otherwise it may require to show with more than one base case. In the recursive step to prove that k +1 is true, first, we need to prove that the statement is correct for all the numbers less than k+1 also.

Let S is a subset of the set of all natural numbers.

1, 2, ….., I ∈ S and I ∈ N and for any k ≥ l, we have 1, 2, . . . , k ∈ S implies k + 1 ∈ S.

Then S = N.

We will show it in terms of a group of propositions.

Let P1, P2, P3, . . . be a group of propositions. Suppose

• Base Case

For some l ∈ N,

P1, P2, . . . , Pl

are all correct, and

• Recursive Step

For any k ≥ l, if P1, P2, . . . , Pk are all correct, then P k+1 is also correct;

Then Pn is true for all n ∈ N.

Check:

• JEE Main Study Notes for Integral Calculus
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Example

Prove that every natural number that is, n > 1 is either prime or a product of prime numbers using the second principle of induction.

Solution:

For each n ≥ 2, let Pn is the set of numbers that either:

• n is prime, or
• n is a product of primes,

n = p1p2 · · · pr, all pi prime.

We shall prove P2, P3, . . . are all true.

In this case we only need one base case and we will start at n = 2

Base Case: 2 is a prime number, so P2 is true.

Recursive Step: Suppose k ≥ 2 is such that P2, P3, . . . , Pk is all true.

Hence all the numbers of 2, 3, . . . , k are either prime or a product of primes.

Now we will show P k+1 is true: that is, k + 1 is prime or a product of primes.

There are two options: either k + 1 is prime or not.

• If k + 1 is prime, then P k+1 is true.
• If k + 1 is not prime, then by definition there exists integers p, q for which k + 1 = pq, where 1 < p, q < k + 1.

But by the induction hypothesis, both Pp and Pq are true.

So we can write

p = p1p2 · · · pr,

q = q1q2 · · · qs,

where p1, p2, . . . , pr, q1, q2, . . . , qs are primes.

So

k + 1 = pq = p1p2 · · · prq1q2 · · · qs,

which is a product of primes. So P k+1 is true.

So whether or not k + 1 is prime, P k+1 is true.

Hence by the second principle of induction, Pn is true for all n ≥ 2.

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JEE Main Sample Questions on Mathematical Induction

Question: If p(n) = n2 + n,∀ n ∈ N, then p(n) is _____________ integer. Justify.

Solution:

P(n) = n2 + n:

P(1) = 1 + 1 = 2;

P(2) = 22 + 2 = 6;

P(3) = 32 + 3 = 12

We observe that ∀n∈N,P(n) is always an even positive integer.

Question: The number (492 − 4) (493 − 49) is divisible by__ !.

Solution:

(492 − 4) (493 − 49) = (492 − 22) × 49 (492 − 1) = 51 * 47 * 49 * 50 * 48 = 51 * 50 * 49 * 48 * 47 is the product of five consecutive integers. Hence, it is divisible by 5!.

Question: What is the least positive remainder when 2301 is divided by 5?

Solution:

24 ≡ 1 (mod 5);

⇒(24)75 ≡ (1)75 (mod 5) i.e.

2300 ≡ 1 (mod 5)

⇒2300 × 2 ≡ (1.2) (mod 5)

⇒2301 ≡ 2 (mod 5),

∴ Least positive remainder is 2.

Preparation Tips for Mathematical Induction

• Memorize as many formulae as you can.
• Class Notes are the perfect example of an initiating step that helps you begin from scratch and build your concepts strong slowly. Pen down all the important information you receive in the class.
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