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Principle of Mathematical Induction

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Mathematical induction is a concept in mathematics that is used to prove various mathematical statements and theorems. The principle of mathematical induction is sometimes referred to as PMI. It is a technique that is used to prove the basic theorems in mathematics which involve the solution up to n finite natural terms. 

The principle of Mathematical Induction is widely used in proving various statements such as a sum of first n natural numbers is given by the formula n(n+1)/2. This can be easily proved using the Principle of Mathematical Induction. In this article, we will learn about the principle of mathematical induction, its statement, its example, and others in detail.

What is Mathematical Induction?

Mathematical Induction is one of the fundamental methods of writing proofs and it is used to prove a given statement about any well-organized set. Generally, it is used for proving results or establishing statements that are formulated in terms of n, where n is a natural number. Suppose P(n) is a statement for n natural number then it can be proved using the Principle of Mathematical Induction, Firstly we will prove for P(1) then let P(k) is true then prove for P(k+1). If P(k+1) holds true then we say that P(n) is true by the principle of mathematical induction.

We can compare mathematical induction to falling dominoes. When a domino falls, it knocks down the next domino in succession. The first domino knocks down the second one, the second one knocks down the third, and so on. In the end, all of the dominoes will be bowled over. But there are some conditions to be fulfilled:

  • The starting domino must fall to set the knocking process in action. This is the base step.
  • The distance between dominoes must be equal for any two adjacent dominoes. Otherwise, a certain domino may fall down without bowling over the next. Then the sequence of reactions will stop. Maintaining the equal inter-domino distance ensures that P(k) ⇒ P(k + 1) for each integer k ≥ a. This is the inductive step.

Principle of Mathematical Induction Statement

Any statement P(n) which is for “n” natural number can be proved using the Principle of Mathematical Induction by following the below steps,

Step 1: Verify if the statement is true for trivial cases (n = 1) i.e. check if P(1) is true.

Step 2: Assume that the statement is true for n = k for some k ≥ 1 i.e. P(k) is true.

Step 3: If the truth of P(k) implies the truth of P(k + 1), then the statement P(n) is true for all n ≥ 1.

The image added below contains all the steps of Mathematical Induction

Principle of Mathematical Induction

 

The first statement is the fact and if it is not possible for all P(n) to hold true at n = 1 then these statements are true for some other values of n say n = 2, n = 3, and others.

If the statement is true for P(k) then if P(k+1) is proven to be true then we say that P(n) is true for all n belonging to Natural Numbers (N)

Mathematical Induction Steps

Various steps used in Mathematical Induction are named accordingly. The names of the various steps used in the principle of mathematical induction are,

  • Base Step: Prove P(k) is true for k =1
  • Assumption Step: Let P(k) is true for all k in N and k > 1
  • Induction Step: Prove P(k+1) is true using basic mathematical properties.

If the above three steps are proved then we can say that “By the principle of mathematical induction, P(n) is true for all n belonging to N”.

Mathematical Induction Example

Mathematical induction is used to prove various statements we can learn this with the help of the following example.

For any positive integer number n, prove that n3 + 2n is always divisible by 3

Solution:

Let P(n): n3 + 2n is divisible by 3 be the given statement.

Step 1: Basic Step

Firstly we prove that P(1) is true. Let n = 1 in n3 + 2n
= 13 + 2(1) 
= 3

As 3 is divisible by 3. Hence, P(1) is true.

Step 2: Assumption Step

Let us assume that P(k) is true

Then, k3 + 2k is divisible by 3

Thus, we can write it as k3 + 2k = 3n,             (where n is any positive integer)….(i)

Step 3: Induction Steps

Now we have to prove that algebraic expression (k + 1)3 + 2(k + 1) is divisible by 3

= (k + 1)3 + 2(k + 1) 

= k3 + 3k2 + 5k + 3

= (k3 + 2 k) + (3k2 + 3k + 3)

from eq(i)

= 3n + 3(k2 + k + 1)

= 3(n + k2 + k + 1)

As it is a multiple of 3 we can say that it is divisible by 3.

Thus, P(k+1) is true i.e. (k + 1)3 + 2(k + 1) is be divisible by 3. Now by the Principle of Mathematical Induction, we can say that, P(n): n3 + 2n is divisible by 3 is true.

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Solved Examples on Mathematical Induction

Example 1: For all n ≥ 1, prove that, 12 + 22 + 32+….+n2 = {n(n + 1) (2n + 1)} / 6

Solution:

Let the given statement be P(n),

P(n):1^2+ 2^2 + 3^2+ \ldots+ n^2 = \frac{n(n + 1) (2n + 1)}{6} \\~\\ \text{For n=1} \\ P(1):\frac{1(1+1)(2×1+1)}{6} = 1

Now, let’s take a positive integer, k, and assume P(k) to be true i.e.,

1^2 + 2^2 + 3^2 +....+k^2 = \frac{k(k+1)(2k+1)}{6}

We shall now prove that P(k + 1) is also true, so now we have,

P(k + 1) = P(k) + (k + 1)2  

= \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \\ \ \\ = \frac {k(k+1)(2k+1)+6{(k+1)}^2}{6} \\ \ \\ =  (k+1) \frac{( 2k^2 + k) + 6(k+1)}{6} \\ \ \\ =\frac{(k+1)(2k^2 +7k+6)}{6} \\ \ \\ =\frac{(k+1) (k+2) (2k+3)}{6}\\ \ \\ =\frac{(k+1) ((k+1)+1) (2(k+1) +1)}{6}

Thus P(k + 1) is true, whenever P(k) is true for all natural numbers. Hence, by the process of mathematical induction, the given result is true for all natural numbers.

Example 2: For all n ≥ 1, prove that, 1.2.3 + 2.3.4 + 3.4.5+…+n(n + 1) (n + 2) = {n (n + 1) (n + 2) ( n + 3)} / 4

Solution: 

Let the given statement be S(n),

S(n):1.2.3+ 2.3.4 + 3.4.5+\ldots+ n.(n+1)(n+2) = \frac{n(n + 1)(n + 2)(n+3)}{4}\\ \ \\ \text{For n=1,} \\ S(1):\frac{1(1+1)(1+2)(1+3)}{4} = 6\\ \text{which is true.}

Now, let’s take a positive integer, k, and assume S(k) to be true i.e.,

S(k):1.2.3+ 2.3.4 + 3.4.5+\ldots+ k.(k+1)(k+2) = \frac{k(k+ 1)(k + 2)(k+3)}{4}

We shall now prove that  S(k + 1) is also true, so now we have,

S(k+1):S(k) + (k+1)(k+2)(k+3)\\ \ \\ \Rightarrow S(k+1): \frac{k(k+ 1)(k + 2)(k+3)}{4} + (k+1)(k+2)(k+3)\\ \ \\ \Rightarrow S(k+1): \frac{k(k+ 1)(k + 2)(k+3)+ \ 4(k+1)(k+2)(k+3)}{4} \\ \ \\ \Rightarrow S(k+1):  \frac{(k+1)(k+2)(k+3)(k+4)}{4}\\ \ \\ \Rightarrow S(k+1): \frac{ (k+1)\{(k+1)+1\}\{(k+1)+2\}\{(k+1)+3\} }{4} \\

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 3: For all n ≥ 1, prove that, 1 + 3 + 5 +… + 2n – 1 = n2

Solution: 

Let the given statement be S(n), 

and S(n) = 1 + 3 + 5+… +2n – 1 = n2

For n = 1, 2 × 1 – 1 = 1 Thus S(1) is true .

Now, let’s take a positive integer, k, and assume S(k) to be true i.e.,

S(k) = 1+ 3 + 5+…+(2k – 1) = k

We shall now prove that  S(k + 1) is also true, so now we have,

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

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

⇒ L.H.S = S(k) + 2k + 1

⇒ L.H.S = k2 + 2k + 1

⇒ L.H.S = (k + 1)

⇒ L.H.S = R.H.S

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 4: For all n ≥ 1, prove that, 1.2 + 2.3 + 3.4 +…+ n(n + 1) = {n(n + 1)(n + 2)} / 3

Solution: 

Let the given statement be S(n),

\ S(n):1.2+ 2.3 + 3.4+ ……+ n.(n+1) = \frac{n(n + 1)(n + 2)}{3}\\ \ \\ \text{for n=1,} \\ \ S(1) : \frac{1(1+1)(1+2)}{3} = 2\\ \text{which is true.} \\

Now, let’s take a positive integer, k, and assume S(k) to be true i.e.,

S(k):1.2+ 2.3 + 3.4+ ……+ k.(k+1) = \frac{k(k+ 1)(k + 2)}{3} \ \\

We shall now prove that S(k + 1) is also true, so now we have,

S(k+1) : S(k) + (k+1)(k+2)\\ \ \\ \Rightarrow S(k+1) : \frac{k(k+ 1)(k + 2)}{3} + (k+1)(k+2)\\ \ \\ \Rightarrow S(k+1) :\frac{k(k+ 1)(k + 2)+ 3(k+1)(k+2)}{3} \\ \ \\ \Rightarrow S(k+1) :\frac{(k+1)(k+2)(k+3)}{3}\\ \ \\ \Rightarrow S(k+1) :\frac{ (k+1)\{(k+1)+1\}\{(k+1)+2\} }{3}

Thus S(k + 1) is true, whenever S(k) is true for all natural numbers. And we initially showed that S(1) is true thus S(n) is true for all natural numbers.

Example 5: Prove an = a1 + (n – 1) d, is the general term of any arithmetic sequence.

Solution: 

For n = 1, we have an = a1 + (1 – 1) d = a1, so the formula is true for n = 1,

Let us assume that the formula ak = a1 + (k – 1) is true for all natural numbers. 

We shall now prove that the formula is also true for k+1, so now we have,

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

We assumed that ak = a1 + (k – 1) d, and by the definition of an arithmetic sequence ak+ 1 – ak = d,

Then, ak + 1 – ak 

= (a1 + k · d) – (a1 + (k – 1)d)
= a1 – a1 + kd – kd + d
= d

Thus the formula is true for k + 1, whenever it is true for k. And we initially showed that the formula is true for n = 1. Thus the formula is true for all natural numbers.

FAQs on Mathematical Induction

Q1: What is the Mathematical Induction Principle?

Answer:

Principle of  Mathematical Induction is a principle which says that for any statement P(n) if its true for any arbitrary value ‘a’ if P(a) is true and if we take P(k) to be true then by proving P(k+1) to be true we can prove that P(n) to be true for all n ≥ a, and n belonging to Natural numbers.

Q2: What is the Use of Mathematical Induction?

Answer:

Mathematical Induction is the basic principle used in mathematics to prove the basic statements in mathematics which can not be easily proved by other means.

Q3: What is the Principle of Mathematical Induction in Matrices?

Answer:

The Principle of Mathematical Induction in Matrices is a basic principle which is used to prove the basic statements in matrices which are not easily proved by other means.

Q4: How to Apply The Principle of Mathematical Induction?

Answer:

The Principle of Mathematical Induction is used to prove mathematical statements suppose we have to prove a statement P(n) then the steps applied are,

Step 1: Prove P(k) is true for k =1

Step 2: Let P(k) is true for all k in N and k > 1

Step 3: Prove P(k+1) is true using basic mathematical properties.

Thus, if P(k+1) is true then we say that P(n) is true.

Q5: What are the Steps to Solve a Problem using Mathematical Induction?

Answer:

The three basic steps used in Mathematical Induction are

  • Base Step
  • Assumption Step
  • Induction Step

Last Updated : 27 May, 2023
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