Principle of Mathematical Induction For Class 11 Notes

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Deductive reasoning is one of the primary basis of mathematical thinking. Inductive reasoning is frequently used in mathematics and is a primary aspect of scientific reasoning, where collecting and analysing data is the norm. The word induction means generalisation from specific cases.

Learn more: Principles of Mathematical Induction with Examples

Principle of Mathematical Induction

There are disciplines in Maths (such as algebra), where the statements are generated in terms of n where n is a positive integer. Hence, such statements are proved by the principle of mathematical induction.

• Each of such statements are represented by P(n) which is linked with n (positive integer)
• Putting n = 1, we have to check the correctness of the statement
• Now, assuming P(k) as true, the truth value of P(k+1) is established, where k is a positive integer

For Example: P(n): 1³ +2³ + 3³ + ….. +n³ = (n(n+1) / 2)² is the true statement for the values of n, where n is a natural number.

Step 1: Put n = 1 and find the value of P(n). P(1) is true.

Step 2. Assume P(k) is true, for the positive integer, k.

Step 3: Now prove, P(k+1) is true whenever P(k) is true

List of Class 11 Maths Notes

Solved Examples

Q.1: Prove that 2ⁿ > n for all positive integers n.

Solution: Let us take P(n): 2ⁿ > n

Now putting, n =1;

2¹>1

Which is true. So, P(1) is true

Let P(k) is true for any positive integer k,

2^k > k … (1)

Now, we need to prove P(k+1) is true whenever P(k) is true.

Multiply by 2 on both sides on eq.(1)

2. 2k > 2k

2^{k + 1} > 2k = k + k > k + 1

Hence, P(k + 1) is true when P(k) is true.

Therefore, P(n) is true for every positive integer n, by principle of mathematical induction.

Q.2: Prove that 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2, by principle of mathematical induction.

Solution: Let us write the given statement as:

P(n): 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

Putting n = 1, we get;

P(1): 1.2 = 2 = (1 – 1) 2¹⁺¹ + 2 = 0 + 2 = 2

The statement P(1) is true.

Let us assume that P(k) is true for a positive integer, k.

1.2 + 2.2² + 3.2² + … + k.2^k = (k – 1) 2^{k + 1} + 2 … (i)

Now, we need to prove that P(k+1) is true, whenever P(k) is true.

P(k+1):

1.2 + 2.2² + 3.2² + … + k.2^k + (k+1)2^k+1

= (k – 1) 2^{k + 1} + 2 + (k+1)2^k+1 [By eq.(i)]

= 2^k+1 [(k-1) + k+1)] + 2

= 2^k+1 .2k + 2

= k2^k+2 + 2

= [(k+1)-1]2^k+2 + 2

Hence, P(k+1) is true whenever P(k) is true.

Thus, by the principle of mathematical induction, statement P (n) is true for all natural numbers (n).

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