Principle of Mathematical Induction Problems

There are some results or statements in algebra which are formulated in terms of n, where n is a positive integer. We use the principle of mathematical induction to prove such statements. In this article, we come across the principle of mathematical induction problems. 

Mathematical induction involves the composition of results or statements. In common, it is used for the set of natural numbers. The statement that needs to be proven is termed a deduction. For example, in the dominoes, if the first domino is knocked, it falls. If any domino is knocked, then the one next to it will also fall.

If the above two steps are satisfied for P(n) natural numbers, then the principle of mathematical induction is valid.

• In some instances, the statement is true for the results n ≥ 7, then the steps will start from n = 7 and hence, check the result for n = 7 and so on.
• The second property given in the principle of mathematical induction is conditional. It doesn’t confirm that the statement is valid for n = k, but if it holds good for n = k, then it is true for n = k + 1 also.

The given statement is true for n=k is a presumption and hence called the hypothesis of induction.

Also Read: Limits Continuity and Differentiability

Solved Problems on the Principle of Mathematical Induction for IIT JEE

Example 1: Prove that 3ⁿ > n for all positive integers n.

Solution:

Let P(n) = 3ⁿ > n

Put n = 1.

⇒ P(1) = 3¹ > 1

So P(1) is true.

Now, assume that P(k) is true for any positive integer k, i.e.,

3^k>k  …(1)

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

Multiply both sides of (1) by 3.

3.3^k> 3k

⇒ 3^k+1 > k + k + k > k + 1

∴ P(k + 1) is true when p(k) is true. 

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

Example 2: Prove that the sum of the first n natural numbers is n(n + 1)/2 through mathematical induction.

Solution: 

Let P(n) = n(n + 1)/2

For n = 1,

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

i.e., P(n) is true for n = 1

For n = 2,

P(2) = 2(2 + 1)/2 = 3

i.e., P(n) is true for n = 2

Let P(n) is true for n = k

That means, P(k) = k(k + 1)/2

Let n = k + 1

P(k + 1) = 1 + 2 + 3 + … + k + (k + 1)

= [k(k + 1)/2] + (k + 1)

= [k(k + 1) + 2(k + 1)]/ 2

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

= (k + 1)[(k + 1) + 1]/2

P(n) is true for n = k + 1

Thus, P(n) is true all n ∈ N

Therefore, by the principle of mathematical induction, the sum of the first n natural numbers is n(n + 1)/2.

Example 3: Prove using mathematical induction that 

Solution:

Let

For n = 1,

P(1) = 1² = [1(1 + 1)(2 × 1 + 1)]/6

1² = (2 × 3)/6

1 = 1

Thus, P(n) is true for n = 1

For n = 2,

P(2) = 1² + 2² = [2(2 + 1)(2 × 2 + 1)]/6

1 + 4 = (2 × 3 × 5)/6

5 = 5

Thus, P(n) is true for n = 2

Let P(n) is true for n = k

So,

Let us assume that P(n) is also true for n = k + 1

1² + 2² + … + k² + (k + 1)² = [k(k + 1)(2k + 1)/6] + (k + 1)²

= [k(k + 1)(2k + 1) + 6(k + 1)²]/ 6

Thus, P(n) is true for n = k + 1

Hence, P(n) is true for all n ∈ N, i.e., for all positive integers.

Therefore, by the principle of mathematical induction,

Example 4: If x ∈ {1, 2 , 3 . . . . 9} and f_n (x) = x x x . . . x (n digits), then what is f ²_n(3) + f_n (2)?

Solution:

f_n(3) = 3 3 3 . . . 3 (n digits) and

f ²_n(3) = 9 9 9 . . . 9 (n digits)

f_n (2) = 2 2 2 . . . 2 (n digits)

∴ f ²_n(3) + f_n (2) = 1 2 . . . 2 2 2 1 ((n+1)digits)

f_n (1) = 1 1 1 . . . 1 (n digits)

∴ f ²_n(3) + f_n (2) = 1 2 2 . . . 2 1 ((n+1)digits)

Example 5: 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.

Also Refer: Binomial Theorem