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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Brief Principle, & Proof of Mathematical Induction

Brief Principle, & Proof of Mathematical Induction

in mathematics is a method that may be applied to demonstrate that a proposition, a formula, or a theorem is true for all natural numbers. This method is known as "mathematical induction."

Table of Content
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Mathematical induction is a method of mathematical proof that may be used to prove a given assertion about any well-organized set. This method can be used for any mathematical problem. In general, it is used for proving conclusions or establishing propositions that are phrased in terms of n, where n is a natural number. This type of proof is known as inductive reasoning. 

Principles

Check to see if the assertion is correct for simple scenarios such as n = a, which would indicate that P(a) is accurate. [Base Case]

  • Assume that the assertion is correct for n = k when k is less than a, which would mean that the condition P(k) holds true. [Hypothesis of Inductive Reasoning]
  • If it is true that P(k) implies that it is also true that P(k + 1), then the statement P(n) is valid for any values of n that are less than or equal to a.
  • The fact that P(k) => P(k + 1) => P(k + 2)…. is true is demonstrated by the combination of the base step and the inductive step. As a result, the statement P(n) is valid for all numbers smaller than a. 

The process of mathematical induction is analogous to that of falling dominoes. When one domino is knocked over, it causes the subsequent dominoes in the line to also fall. There is no way around it. However, there are a few prerequisites that need to be met:

  • The process of knocking dominoes over can only begin once the initial domino has been knocked over. This is the fundamental first step
  • Each pair of dominoes that are adjacent to one another must have the same spacing between them. In that case, the fall of one domino might not result in the collapse of the subsequent one. After that, the chain of reactions will come to an end. Keeping the distance between each domino the same assures that P(k) is less than P(k + 1) for every integer k that is less than a. This is the first phase in the inductive process

Using the Principle of Mathematical Induction

The most important application of the Principle of Mathematical Induction is to demonstrate the validity of assertions of the type.

(∀n∈N)(P(n)) .

where P(n) is a free-form statement of some kind. It is important to keep in mind that a universally quantified assertion, such as the one that came before it, is considered to be true if and only if the truth set T of the open phrase P(n) is the set N. Since the conclusion of the Principle of Mathematical Induction is that T=N must be demonstrated, this will be our primary focus. We need to carry out this step in order to demonstrate that the hypothesis of the mathematical induction principle is accurate.

Demonstrate that 1 = T. To put it another way, show that P(1) is correct.

Show that if k is smaller than T, then (k+1) must be smaller. To put it another way, demonstrate that if P(k) is true, then P(k+1) is also true.

Conclusion

A proof that is based on induction requires two cases. The first scenario, often known as the base case or foundation, validates the assertion that n = 0 without presupposing any prior familiarity with the other scenarios. The induction step, which is the second instance, demonstrates that if the statement is true for any particular situation, such as n = k, then it must also be true for the subsequent case, which is n = k + 1. After completing these two stages, we have proven that the assertion is valid for all natural numbers n. It is not always the case that the base case starts with n = 0, but more often than not, it starts with n = 1, and it might start with any fixed natural number n = N. This establishes that the assertion is true for all natural numbers that are less than or equal to N.

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Convince us an = a1 + (n – 1) d, is the general term of any arithmetic sequence of a mathematical expression.

Answer. Because a = a1 + (1 – 1) d = a1 corresponds to a1 when n = 1, proving that the formula is accurate when n = 1. Let...Read full

Prove that, 1 + 3 + 5 … 2n – 1 = n2 for n>=1

Answer.Let the supplied statement be S (n), and S(n) = 1 + 3 + 5 … 2n...Read full

What does mathematical induction imply?

Answer. The term “mathematical induction” refers to a method for determining natural number outcomes. Th...Read full

What exactly does it mean to infer anything using mathematics?

Answer. A procedure that is used to produce findings for the natural numbers is referred to as mathematical inductio...Read full

Answer. Because a = a1 + (1 – 1) d = a1 corresponds to a1 when n = 1, proving that the formula is accurate when n = 1.

Let us make the assumption that the formula ak = a1 + (k – 1) describes all of the natural numbers correctly.

Next, we will demonstrate that the formula also holds true for k plus one, and after we do so, we will obtain

ak + 1 = a1 + [(k + 1) – 1] d = a1 + k · d.

We made the assumption that ak = a1 + (k – 1) d, and according to the definition of an arithmetic sequence as ak+ 1 – ak = d, we found that our assumption was correct.

therefore, ak + 1 – ak

= (a1 + k · d) – (a1 + (k – 1)d)

= a1 – a1 + kd – kd + d

= d

Therefore, anytime the formula is true for k, it also holds true for k + 1. In the beginning, we demonstrated that the formula is accurate when n is equal to 1. As a result, the formula is valid for all of the natural numbers.

Answer.Let the supplied statement be S (n),

and S(n) = 1 + 3 + 5 … 2n – 1 = n2

Since n equals one, 2 times one minus one equals 12, proving that S(1) is correct.

Now, let’s choose a positive integer and suppose that the statement S(k) is correct, which means:

S(k) = 1+ 3 + 5 .. (2k – 1) = k²

Next, we will demonstrate that S(k + 1) is likewise accurate, and at this point, we will have,

1 + 3 + 5 .. (2(k + 1) – 1) = (k + 1)²

L.H.S: 1 + 3 + 5 + …. (2k – 1) + 2k + 2 – 1

= S(k) + 2k + 1

= k2 + 2k + 1

= (k + 1)2

= R.H.S

Therefore, anytime the statement S(k) is true for all natural numbers, the statement S(k + 1) also holds true. And to begin, we demonstrated that S(1) is correct, which means that S(n) is correct for all natural integers.

Answer. The term “mathematical induction” refers to a method for determining natural number outcomes. This approach is typically used to demonstrate that a statement or theorem is true for all natural numbers.

Answer. A procedure that is used to produce findings for the natural numbers is referred to as mathematical induction, and it is defined as such. In most cases, this approach is used to demonstrate that the assertion or theorem is valid for all natural number values.

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