RD Sharma Solutions for Class 11 Chapter 12 - Mathematical Induction Exercise 12.1

In Exercise 12.1 of Chapter 12, we shall discuss problems based on the first principle of mathematical induction. Students can refer to RD Sharma Class 11 Maths as the reference material to score high marks and come out with flying colours in their board exams. Keeping in mind the concepts, subject experts have prepared the solutions in accordance with the latest CBSE syllabus. The solutions are solved in a step-by-step format using shortcut methods so that every student can easily remember the steps. Students can easily download the RD Sharma Class 11 Solutions for Maths, which is available in PDF format from the links given below.

RD Sharma Solutions for Class 11 Maths Exercise 12.1 Chapter 12 – Mathematical Induction

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Exercise 12.2 Solutions

Access answers to RD Sharma Solutions for Class 11 Maths Exercise 12.1 Chapter 12 – Mathematical Induction

1. If P (n) is the statement “n (n + 1) is even”, then what is P (3)?
Solution:

Given:

P (n) = n (n + 1) is even.

So,

P (3) = 3 (3 + 1)

= 3 (4)

= 12

Hence, P (3) = 12, and P (3) is also even.

2. If P (n) is the statement “n³ + n is divisible by 3”, prove that P (3) is true but P (4) is not true.

Solution:

Given:

P (n) = n³ + n is divisible by 3

We have P (n) = n³ + n

So,

P (3) = 3³ + 3

= 27 + 3

= 30

P (3) = 30, So it is divisible by 3

Now, let’s check with P (4)

P (4) = 4³ + 4

= 64 + 4

= 68

P (4) = 68, so it is not divisible by 3

Hence, P (3) is true, and P (4) is not true.

3. If P (n) is the statement “2ⁿ ≥ 3n”, and if P (r) is true, prove that P (r + 1) is true.

Solution:

Given:

P (n) = “2ⁿ ≥ 3n” and p(r) is true.

We have P (n) = 2ⁿ ≥ 3n

Since P (r) is true

So,

2^r≥ 3r

Now, let’s multiply both sides by 2

2×2^r≥ 3r×2

2^{r + 1}≥ 6r

2^{r + 1}≥ 3r + 3r [since 3r>3 = 3r + 3r≥3 + 3r]

∴ 2^{r + 1}≥ 3(r + 1)

Hence, P (r + 1) is true.

4. If P (n) is the statement “n² + n” is even”, and if P (r) is true, then P (r + 1) is true
Solution:

Given:

P (n) = n² + n is even, and P (r) is true, then r² + r is even

Let us consider r² + r = 2k … (i)

Now, (r + 1)² + (r + 1)

r² + 1 + 2r + r + 1

(r² + r) + 2r + 2

2k + 2r + 2 [from equation (i)]

2(k + r + 1)

2μ

∴ (r + 1)² + (r + 1) is Even.

Hence, P (r + 1) is true.

5. Given an example of a statement P (n) such that it is true for all n ϵ N.

Solution:

Let us consider

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

So,

P (n) is true for all natural numbers.

Hence, P (n) is true for all n ∈ N.

6. If P (n) is the statement “n² – n + 41 is prime”, prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
Solution:

Given:

P(n) = n² – n + 41 is prime.

P(n) = n² – n + 41

P (1) = 1 – 1 + 41

= 41

P (1) is Prime.

Similarly,

P(2) = 2² – 2 + 41

= 4 – 2 + 41

= 43

P (2) is prime.

Similarly,

P (3) = 3² – 3 + 41

= 9 – 3 + 41

= 47

P (3) is prime

Now,

P (41) = (41)² – 41 + 41

= 1681

P (41) is not prime

Hence, P (1), P(2), and P (3) are true, but P (41) is not true.