RD Sharma Class 11 Solutions Chapter 12 Exercise 12.2 (Updated for 2023-24)

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RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2

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Prove the following by the principle of mathematical induction.

1. 1 + 2 + 3 + … + n = n (n +1)/2, i.e., the sum of the first n natural numbers is n (n + 1)/2.

Solution:

Let us consider P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2

For, n = 1

LHS of P (n) = 1

RHS of P (n) = 1 (1+1)/2 = 1

So, LHS = RHS

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

Let us consider P (n) to be true for n = k, so

1 + 2 + 3 + …. + k = k (k+1)/2 … (i)

Now,

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

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

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

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

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

P (n) is true for all n ∈ N

So, by the principle of Mathematical Induction,

P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2 is true for all n ∈ N.

2. 1² + 2² + 3² + … + n² = [n (n+1) (2n+1)]/6

Solution:

Let us consider P (n) = 1² + 2² + 3² + … + n² = [n (n+1) (2n+1)]/6

For, n = 1

P (1) = [1 (1+1) (2+1)]/6

1 = 1

P (n) is true for n = 1

Let P (n) is true for n = k, so

P (k): 1² + 2² + 3² + … + k² = [k (k+1) (2k+1)]/6

Let’s check for P (n) = k + 1, so

P (k) = 1² + 2² + 3² + – – – – – + k² + (k + 1)² = [k + 1 (k+2) (2k+3)] /6

= 1² + 2² + 3² + – – – – – + k² + (k + 1)²

= [k + 1 (k+2) (2k+3)] /6 + (k + 1)²

= (k +1) [(2k² + k)/6 + (k + 1)/1]

= (k +1) [2k² + k + 6k + 6]/6

= (k +1) [2k² + 7k + 6]/6

= (k +1) [2k² + 4k + 3k + 6]/6

= (k +1) [2k(k + 2) + 3(k + 2)]/6

= [(k +1) (2k + 3) (k + 2)] / 6

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

Hence, P (n) is true for all n ∈ N.

3. 1 + 3 + 3² + … + 3^n-1 = (3ⁿ – 1)/2

Solution:

Let P (n) = 1 + 3 + 3² + – – – – + 3^{n – 1} = (3ⁿ – 1)/2

Now, for n = 1

P (1) = 1 = (3¹ – 1)/2 = 2/2 =1

P (n) is true for n = 1

Now, let’s check for P (n) is true for n = k

P (k) = 1 + 3 + 3² + – – – – + 3^{k – 1} = (3^k – 1)/2 … (i)

Now, we have to show that P (n) is true for n = k + 1

P (k + 1) = 1 + 3 + 3² + – – – – + 3^k = (3^k+1 – 1)/2

Then, {1 + 3 + 3² + – – – – + 3^{k – 1}} + 3^{k + 1 – 1}

= (3k – 1)/2 + 3^k using equation (i)

= (3k – 1 + 2×3^k)/2

= (3×3 k – 1)/2

= (3^k+1 – 1)/2

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

Hence, P (n) is true for all n ∈ N.

4. 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)

Solution:

Let P (n) = 1/1.2 + 1/2.3 + 1/3.4 + … + 1/n(n+1) = n/(n+1)

For, n = 1

P (n) = 1/1.2 = 1/1+1

1/2 = 1/2

P (n) is true for n = 1

Let’s check for P (n) is true for n = k,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k+1) + k/(k+1) (k+2) = (k+1)/(k+2)

Then,

1/1.2 + 1/2.3 + 1/3.4 + … + 1/k(k+1) + k/(k+1) (k+2)

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

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

= 1/(k+1) [k² + 2k + 1]/(k+2)

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

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

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

Hence, P (n) is true for all n ∈ N.

5. 1 + 3 + 5 + … + (2n – 1) = n², i.e., the sum of the first n odd natural numbers is n².

Solution:

Let P (n): 1 + 3 + 5 + … + (2n – 1) = n²

Let us check if P (n) is true for n = 1

P (1) = 1 =1²

1 = 1

P (n) is true for n = 1

Now, Let’s check that P (n) is true for n = k

P (k) = 1 + 3 + 5 + … + (2k – 1) = k² … (i)

We have to show that

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

Now,

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

= k² + (2k + 1)

= k² + 2k + 1

= (k + 1)²

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

Hence, P (n) is true for all n ∈ N.

6. 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1) (3n+2) = n/(6n+4)

Solution:

Let P (n) = 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3n-1) (3n+2) = n/(6n+4)

Let us check if P (n) is true for n = 1

P (1): 1/2.5 = 1/6.1+4 => 1/10 = 1/10

P (1) is true.

Now,

Let us check that P (k) is true and prove that P (k + 1) is true.

P (k): 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k-1) (3k+2) = k/(6k+4)

P (k +1): 1/2.5 + 1/5.8 + 1/8.11 + … + 1/(3k-1)(3k+2) + 1/(3k+3-1)(3k+3+2)

: k/(6k+4) + 1/(3k+2)(3k+5)

: [k(3k+5)+2] / [2(3k+2)(3k+5)]

: (k+1) / (6(k+1)+4)

P (k + 1) is true.

Hence proved by mathematical induction.

7. 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n-2)(3n+1) = n/3n+1

Solution:

Let P (n) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n-2)(3n+1) = n/3n+1

Let us check for n = 1

P (1): 1/1.4 = 1/4

1/4 = 1/4

P (n) is true for n = 1.

Now, let us check for P (n) is true for n = k and prove that P (k + 1) is true.

P (k) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3k-2)(3k+1) = k/3k+1 … (i)

So,

[1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3k-2)(3k+1)]+ 1/(3k+1)(3k+4)

= k/(3k+1) + 1/(3k+1)(3k+4)

= 1/(3k+1) [k/1 + 1/(3k+4)]

= 1/(3k+1) [k(3k+4)+1]/(3k+4)

= 1/(3k+1) [3k² + 4k + 1]/ (3k+4)

= 1/(3k+1) [3k² + 3k+k+1]/(3k+4)

= [3k(k+1) + (k+1)] / [(3k+4) (3k+1)]

= [(3k+1)(k+1)] / [(3k+4) (3k+1)]

= (k+1) / (3k+4)

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

Hence, P (n) is true for all n ∈ N.

8. 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n+1)(2n+3) = n/3(2n+3)

Solution:

Let P (n) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n+1)(2n+3) = n/3(2n+3)

Let us check for n = 1,

P (1): 1/3.5 = 1/3(2.1+3)

: 1/15 = 1/15

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) = k/3(2k+3) … (i)

So,

1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) + 1/[2(k+1)+1][2(k+1)+3]

1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) + 1/(2k+3)(2k+5)

Now substituting the value of P (k), we get

= k/3(2k+3) + 1/(2k+3)(2k+5)

= [k(2k+5)+3] / [3(2k+3)(2k+5)]

= (k+1) / [3(2(k+1)+3)]

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

Hence, P (n) is true for all n ∈ N.

9. 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n-1)(4n+3) = n/3(4n+3)

Solution:

Let P (n) = 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n-1)(4n+3) = n/3(4n+3)

Let us check for n = 1,

P (1): 1/3.7 = 1/(4.1-1)(4+3)

: 1/21 = 1/21

P (n) is true for n =1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) = k/3(4k+3) …. (i)

So,

1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) + 1/(4k+3)(4k+7)

Substituting the value of P (k), we get

= k/(4k+3) + 1/(4k+3)(4k+7)

= 1/(4k+3) [k(4k+7)+3] / [3(4k+7)]

= 1/(4k+3) [4k² + 7k +3]/ [3(4k+7)]

= 1/(4k+3) [4k² + 3k+4k+3] / [3(4k+7)]

= 1/(4k+3) [4k(k+1)+3(k+1)]/ [3(4k+7)]

= 1/(4k+3) [(4k+3)(k+1)] / [3(4k+7)]

= (k+1) / [3(4k+7)]

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

Hence, P (n) is true for all n ∈ N.

10. 1.2 + 2.2² + 3.2³ + … + n.2ⁿ = (n–1) 2^{n + 1} + 2

Solution:

Let P (n) = 1.2 + 2.2² + 3.2³ + … + n.2ⁿ = (n–1) 2^{n + 1} + 2

Let us check for n = 1,

P (1):1.2 = 0.2⁰ + 2

: 2 = 2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.2 + 2.2² + 3.2³ + … + k.2^k = (k–1) 2^{k + 1} + 2 …. (i)

So,

{1.2 + 2.2² + 3.2³ + … + k.2^k} + (k + 1)2^{k + 1}

Now, substituting the value of P (k), we get

= [(k – 1)2^{k + 1} + 2] + (k + 1)2^{k + 1} using equation (i)

= (k – 1)2^{k + 1} + 2 + (k + 1)2^{k + 1}

= 2^{k + 1}(k – 1 + k + 1) + 2

= 2^{k + 1} × 2k + 2

= k × 2^{k + 2} + 2

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

Hence, P (n) is true for all n ∈ N.

11. 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)

Solution:

Let P (n) = 2 + 5 + 8 + 11 + … + (3n – 1) = 1/2 n (3n + 1)

Let us check for n = 1,

P (1): 2 = 1/2 × 1 × 4

: 2 = 2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k) = 2 + 5 + 8 + 11 + … + (3k – 1) = 1/2 k (3k + 1) … (i)

So,

2 + 5 + 8 + 11 + … + (3k – 1) + (3k + 2)

Now, substituting the value of P (k), we get,

= 1/2 × k (3k + 1) + (3k + 2) by using equation (i)

= [3k² + k + 2 (3k + 2)] / 2

= [3k² + k + 6k + 2] / 2

= [3k² + 7k + 2] / 2

= [3k² + 4k + 3k + 2] / 2

= [3k (k+1) + 4(k+1)] / 2

= [(k+1) (3k+4)] /2

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

Hence, P (n) is true for all n ∈ N.

12. 1.3 + 2.4 + 3.5 + … + n. (n+2) = 1/6 n (n+1) (2n+7)

Solution:

Let P (n): 1.3 + 2.4 + 3.5 + … + n. (n+2) = 1/6 n (n+1) (2n+7)

Let us check for n = 1

P (1): 1.3 = 1/6 × 1 × 2 × 9

: 3 = 3

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.3 + 2.4 + 3.5 + … + k. (k+2) = 1/6 k (k+1) (2k+7) … (i)

So,

1.3 + 2.4 + 3.5 + … + k. (k+2) + (k+1) (k+3)

Now, substituting the value of P (k), we get

= 1/6 k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)

= (k+1) [{k(2k+7)/6} + {(k+3)/1}]

= (k+1) [(2k² + 7k + 6k + 18)] / 6

= (k+1) [2k² + 13k + 18] / 6

= (k+1) [2k² + 9k + 4k + 18] / 6

= (k+1) [2k(k+2) + 9(k+2)] / 6

= (k+1) [(2k+9) (k+2)] / 6

= 1/6 (k+1) (k+2) (2k+9)

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

Hence, P (n) is true for all n ∈ N.

13. 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n² + 6n – 1)/3

Solution:

Let P (n): 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = n(4n² + 6n – 1)/3

Let us check for n = 1,

P (1): (2.1 – 1) (2.1 + 1) = 1(4.1² + 6.1 -1)/3

: 1×3 = 1(4+6-1)/3

: 3 = 3

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) = k(4k² + 6k – 1)/3 … (i)

So,

1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)

Now, substituting the value of P (k), we get

= k(4k² + 6k – 1)/3 + (2k + 1) (2k + 3) by using equation (i)

= [k(4k² + 6k-1) + 3 (4k² + 6k + 2k + 3)] / 3

= [4k³ + 6k² – k + 12k² + 18k + 6k + 9] /3

= [4k³ + 18k² + 23k + 9] /3

= [4k³ + 4k² + 14k² + 14k +9k + 9] /3

= [(k+1) (4k² + 8k +4 + 6k + 6 – 1)] / 3

= [(k+1) 4[(k+1)² + 6(k+1) -1]] /3

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

Hence, P (n) is true for all n ∈ N.

14. 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3

Solution:

Let P (n): 1.2 + 2.3 + 3.4 + … + n(n+1) = [n (n+1) (n+2)] / 3

Let us check for n = 1,

P (1): 1(1+1) = [1(1+1) (1+2)] /3

: 2 = 2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1.2 + 2.3 + 3.4 + … + k(k+1) = [k (k+1) (k+2)] / 3 … (i)

So,

1.2 + 2.3 + 3.4 + … + k(k+1) + (k+1) (k+2)

Now, substituting the value of P (k), we get

= [k (k+1) (k+2)] / 3 + (k+1) (k+2) by using equation (i)

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

= [(k+1) (k+2) (k+3)] /3

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

Hence, P (n) is true for all n ∈ N.

15. 1/2 + 1/4 + 1/8 + … + 1/2ⁿ = 1 – 1/2ⁿ

Solution:

Let P (n): 1/2 + 1/4 + 1/8 + … + 1/2ⁿ = 1 – 1/2ⁿ

Let us check for n = 1,

P (1): 1/2¹ = 1 – 1/2¹

: 1/2 = 1/2

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

Let P (k): 1/2 + 1/4 + 1/8 + … + 1/2^k = 1 – 1/2^k … (i)

So,

1/2 + 1/4 + 1/8 + … + 1/2^k + 1/2^k+1

Now, substituting the value of P (k), we get

= 1 – 1/2^k + 1/2^k+1 by using equation (i)

= 1 – ((2-1)/2^k+1)

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

Hence, P (n) is true for all n ∈ N.

16. 1² + 3² + 5² + … + (2n – 1)² = 1/3 n (4n² – 1)

Solution:

Let P (n): 1² + 3² + 5² + … + (2n – 1)² = 1/3 n (4n² – 1)

Let us check for n = 1,

P (1): (2.1 – 1)² = 1/3 × 1 × (4 – 1)

: 1 = 1

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 1² + 3² + 5² + … + (2k – 1)² = 1/3 k (4k² – 1) … (i)

So,

1² + 3² + 5² + … + (2k – 1)² + (2k + 1)²

Now, substituting the value of P (k), we get,

= 1/3 k (4k² – 1) + (2k + 1)² by using equation (i)

= 1/3 k (2k + 1) (2k – 1) + (2k + 1)²

= (2k + 1) [{k(2k-1)/3} + (2k+1)]

= (2k + 1) [2k² – k + 3(2k+1)] / 3

= (2k + 1) [2k² – k + 6k + 3] / 3

= [(2k+1) 2k² + 5k + 3] /3

= [(2k+1) (2k(k+1)) + 3 (k+1)] /3

= [(2k+1) (2k+3) (k+1)] /3

= (k+1)/3 [4k² + 6k + 2k + 3]

= (k+1)/3 [4k² + 8k – 1]

= (k+1)/3 [4(k+1)² – 1]

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

Hence, P (n) is true for all n ∈ N.

17. a + ar + ar² + … + ar^{n – 1} = a [(rⁿ – 1)/(r – 1)], r ≠ 1

Solution:

Let P (n): a + ar + ar² + … + ar^{n – 1} = a [(rⁿ – 1)/(r – 1)]

Let us check for n = 1,

P (1): a = a (r¹ – 1)/(r-1)

: a = a

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): a + ar + ar² + … + ar^{k – 1} = a [(r^k – 1)/(r – 1)] … (i)

So,

a + ar + ar² + … + ar^{k – 1} + ar^k

Now, substituting the value of P (k), we get,

= a [(r^k – 1)/(r – 1)] + ar^k by using equation (i)

= a[r^k – 1 + r^k(r-1)] / (r-1)

= a[r^k – 1 + r^k+1 – r^‑k] / (r-1)

= a[r^k+1 – 1] / (r-1)

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

Hence, P (n) is true for all n ∈ N.

18. a + (a + d) + (a + 2d) + … + (a + (n-1)d) = n/2 [2a + (n-1)d]

Solution:

Let P (n): a + (a + d) + (a + 2d) + … + (a + (n-1)d) = n/2 [2a + (n-1)d]

Let us check for n = 1,

P (1): a = ½ [2a + (1-1)d]

: a = a

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): a + (a + d) + (a + 2d) + … + (a + (k-1)d) = k/2 [2a + (k-1)d] … (i)

So,

a + (a + d) + (a + 2d) + … + (a + (k-1)d) + (a + (k)d)

Now, substituting the value of P (k), we get,

= k/2 [2a + (k-1)d] + (a + kd) by using equation (i)

= [2ka + k(k-1)d + 2(a+kd)] / 2

= [2ka + k²d – kd + 2a + 2kd] / 2

= [2ka + 2a + k²d + kd] / 2

= [2a(k+1) + d(k² + k)] / 2

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

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

Hence, P (n) is true for all n ∈ N.

19. 5²ⁿ – 1 is divisible by 24 for all n ϵ N

Solution:

Let P (n): 5²ⁿ – 1 is divisible by 24

Let us check for n = 1,

P (1): 5² – 1 = 25 – 1 = 24

P (n) is true for n = 1. Where P (n) is divisible by 24

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 5^2k – 1 is divisible by 24

: 5^2k – 1 = 24λ … (i)

We have to prove,

5^{2k + 1} – 1 is divisible by 24

5^{2(k + 1)} – 1 = 24μ

So,

= 5^{2(k + 1)} – 1

= 5^2k.5² – 1

= 25.5^2k – 1

= 25.(24λ + 1) – 1 by using equation (1)

= 25.24λ + 24

= 24λ

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

Hence, P (n) is true for all n ∈ N.

20. 3²ⁿ + 7 is divisible by 8 for all n ϵ N

Solution:

Let P (n): 3²ⁿ + 7 is divisible by 8

Let us check for n = 1,

P (1): 3² + 7 = 9 + 7 = 16

P (n) is true for n = 1. Where P (n) is divisible by 8

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 3^2k + 7 is divisible by 8

: 3^2k + 7 = 8λ

: 3^2k = 8λ – 7 … (i)

We have to prove,

3^{2(k + 1)} + 7 is divisible by 8

3^{2k + 2} + 7 = 8μ

So,

= 3^{2(k + 1)} + 7

= 3^2k.3² + 7

= 9.3^2k + 7

= 9.(8λ – 7) + 7 by using equation (i)

= 72λ – 63 + 7

= 72λ – 56

= 8(9λ – 7)

= 8μ

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

Hence, P (n) is true for all n ∈ N.

21. 5^{2n + 2} – 24n – 25 is divisible by 576 for all n ϵ N

Solution:

Let P (n): 5^{2n + 2} – 24n – 25 is divisible by 576.

Let us check for n = 1,

P (1): 5^2.1+2 – 24.1 – 25

: 625 – 49

: 576

P (n) is true for n = 1. Where P (n) is divisible by 576

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 5^{2k + 2} – 24k – 25 is divisible by 576

: 5^{2k + 2} – 24k – 25 = 576λ …. (i)

We have to prove,

5^{2k + 4} – 24(k + 1) – 25 is divisible by 576

5^{(2k + 2) + 2} – 24(k + 1) – 25 = 576μ

So,

= 5^{(2k + 2) + 2} – 24(k + 1) – 25

= 5^{(2k + 2)}.5² – 24k – 24– 25

= (576λ + 24k + 25)25 – 24k– 49 by using equation (i)

= 25. 576λ + 576k + 576

= 576(25λ + k + 1)

= 576μ

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

Hence, P (n) is true for all n ∈ N.

22. 3^{2n + 2} – 8n – 9 is divisible by 8 for all n ϵ N

Solution:

Let P (n): 3^{2n + 2} – 8n – 9 is divisible by 8

Let us check for n = 1,

P (1): 3^{2.1 + 2} – 8.1 – 9

: 81 – 17

: 64

P (n) is true for n = 1. Where P (n) is divisible by 8

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 3^{2k + 2} – 8k – 9 is divisible by 8

: 3^{2k + 2} – 8k – 9 = 8λ … (i)

We have to prove,

3^{2k + 4} – 8(k + 1) – 9 is divisible by 8

3^{(2k + 2) + 2} – 8(k + 1) – 9 = 8μ

So,

= 3^{2(k + 1)}.3² – 8(k + 1) – 9

= (8λ + 8k + 9)9 – 8k – 8 – 9

= 72λ + 72k + 81 – 8k – 17 using equation (1)

= 72λ + 64k + 64

= 8(9λ + 8k + 8)

= 8μ

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

Hence, P (n) is true for all n ∈ N.

23. (ab) ⁿ = aⁿ bⁿ for all n ϵ N

Solution:

Let P (n): (ab) ⁿ = aⁿ bⁿ

Let us check for n = 1,

P (1): (ab) ¹ = a¹ b¹

: ab = ab

P (n) is true for n = 1.

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): (ab) ^k = a^k b^k … (i)

We have to prove,

(ab) ^{k + 1} = a^{k + 1}.b^{k + 1}

So,

= (ab) ^{k + 1}

= (ab) ^k (ab)

= (a^k b^k) (ab) using equation (1)

= (a^{k + 1}) (b^{k + 1})

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

Hence, P (n) is true for all n ∈ N.

24. n (n + 1) (n + 5) is a multiple of 3 for all n ϵ N.

Solution:

Let P (n): n (n + 1) (n + 5) is a multiple of 3

Let us check for n = 1,

P (1): 1 (1 + 1) (1 + 5)

: 2 × 6

: 12

P (n) is true for n = 1. Where P (n) is a multiple of 3

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): k (k + 1) (k + 5) is a multiple of 3

: k(k + 1) (k + 5) = 3λ … (i)

We have to prove,

(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3

(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3μ

So,

= (k + 1) [(k + 1) + 1] [(k + 1) + 5]

= (k + 1) (k + 2) [(k + 1) + 5]

= [k (k + 1) + 2(k + 1)] [(k + 5) + 1]

= k (k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)

= 3λ + k² + k + 2(k² + 6k + 5) + 2k + 2

= 3λ + k² + k + 2k² + 12k + 10 + 2k + 2

= 3λ + 3k² + 15k + 12

= 3(λ + k² + 5k + 4)

= 3μ

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

Hence, P (n) is true for all n ∈ N.

25. 7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

Solution:

Let P (n): 7²ⁿ + 2^{3n – 3}. 3n – 1 is divisible by 25

Let us check for n = 1,

P (1): 7² + 2⁰.3⁰

: 49 + 1

: 50

P (n) is true for n = 1. Where P (n) is divisible by 25

Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.

P (k): 7^2k + 2^{3k – 3}. 3k – 1 is divisible by 25

: 7^2k + 2^{3k – 3}. 3^{k – 1} = 25λ … (i)

We have to prove that:

7^{2k + 1} + 2^3k. 3^k is divisible by 25

7^{2k + 2} + 2^3k. 3^k = 25μ

So,

= 7^{2(k + 1)} + 2^3k. 3^k

= 7^2k.7¹ + 2^3k. 3^k

= (25λ – 2^{3k – 3}. 3^{k – 1}) 49 + 2^3k. 3k by using equation (i)

= 25λ. 49 – 2^3k/8. 3^k/3. 49 + 2^3k. 3^k

= 24×25×49λ – 2^3k . 3^k . 49 + 24 . 2^3k.3^k

= 24×25×49λ – 25 . 2^3k. 3^k

= 25(24 . 49λ – 2^3k. 3^k)

= 25μ

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

Hence, P (n) is true for all n ∈ N.

This is the complete blog on the RD Sharma Solutions For Class 11 Maths Chapter 12 Exercise 12.2. To Know more about the CBSE Class 11 Maths exam, as in the comments. 

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