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# Class 11 RD Sharma Solutions – Chapter 12 Mathematical Induction – Exercise 12.2 | Set 1

• Last Updated : 20 May, 2021

### Question 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, P (n) = 1 + 2 + 3 + ….. + n = n (n +1)/2.

Step 1:

Now, let us check P(n) for n = 1.

So, P (1) = 1 (1 + 1)/2 = 1.

So, P(1) is true.

Step 2:

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

Then, 1 + 2 + 3 + …. + k = k (k + 1)/2           … (i)

Step 3:

Now, we have to prove for P(k+1) is true.

(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

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI) P (n) is true for all n ∈ N.

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

Solution:

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

Step 1:

Now, let us check P(n) for n = 1.

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

or, 1 = 1.

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k + 1) is true.

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

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 3. 1 + 3 + 32 + … + 3n – 1 = (3n – 1)/2

Solution:

Let P (n) = 1 + 3 + 32 + ……. + 3n – 1 = (3n – 1)/2

Step 1:

Now, let us check P(n) for n=1.

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

So, P(1) is true.

Step 2:

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

P (k) = 1 + 3 + 32 + ……. + 3k – 1 = (3k – 1)/2       … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

Then,

{1 + 3 + 32 + ……. + 3k – 1} + 3k + 1 – 1

Using equation (i), we get

= (3k – 1)/2 + 3k

= (3k – 1 + 2 × 3k)/2

= (3 × 3 k – 1)/2

= (3k + 1 – 1)/2

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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)

Step 1:

Now, let us check P(n) for n=1.

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

1/2 = 1/2

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k + 1) is true.

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)

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

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

Solution:

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

Step 1:

Now, let us check P(n) for n=1.

P (1) = 1 = 12 = 1

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k + 1) is true.

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

= k2 + (2k + 1)

= k2 + 2k + 1

= (k + 1)2

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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)

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 1/2.5 = 1/(6.1 + 4)

or, 1/10 = 1/10

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

Then, 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)

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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

Step 1:

Now, let us check P(n) for n = 1.

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

Or, 1/4 = 1/4

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

Then, [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)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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)

Step 1:

Now, let us check P(n) for n = 1.

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

or, 1/15 = 1/15

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k + 1) is true.

Then, 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 putting the value of (i), 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)]

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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)

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 1/(3.7) = 1/((4.1 – 1)(4 + 3))

Or, 1/21 = 1/21

So, P(1) is true.

Step 2:

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

P (k) = 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k – 1)(4k + 3) = k/3(4k + 3) … (i)

Step 3:

Now, we have to prove for P(k + 1) is true.

Then, 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k – 1)(4k + 3) + 1/(4k + 3)(4k + 7)

Now putting the value of (i), 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)]

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 10. 1.2 + 2.22 + 3.23 + … + n.2n = (n – 1) 2n+1 + 2

Solution:

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

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 1.2 = 0.20 + 2

or, 2 = 2

So, P(1) is true.

Step 2:

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

P (k) =1.2 + 2.22 + 3.23 + … + k.2k = (k – 1) 2k+1 + 2  … (i)

Step 3:

Now, we have to prove for P(k+1) is true.

Then, [1.2 + 2.22 + 3.23 + … + k.2k] + (k + 1)2k + 1

Now, putting the value of (i), we get,

= [(k – 1)2k+1 + 2] + (k + 1)2k+1

= (k – 1)2k+1 + 2 + (k + 1)2k+1

= 2k+1(k – 1 + k + 1) + 2

= 2k+1 × 2k + 2

= k.2k+2 + 2

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

i.e, P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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)

Step 1:

Now, let us check P(n) for n = 1.

P (1) = 2 = 1/2 × 1 × 4

or, 2 = 2

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

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

Now, putting the value (i), we get,

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

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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)

Step 1:

Now, let us check P(n) for n=1.

P (1) = 1.3 = 1/6 × 1 × 2 × 9

Or, 3 = 3

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

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

Now, putting the value (i), we get,

= 1/6 k (k + 1) (2k + 7) + (k + 1) (k + 3)            (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)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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

Step 1:

Now, let us check P(n) for n = 1.

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

or, 1 × 3 = 1(4 + 6 – 1)/3

or, 3 = 3

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

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

Now, putting the value (i), we get,

= k(4k2 + 6k – 1)/3 + (2k + 1) (2k + 3) (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

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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

Step 1:

Now, let us check P(n) for n = 1.

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

or, 2 = 2

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k + 1) is true.

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

Now, putting the value (i), we get,

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

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

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

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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

Step 1:

Now, let us check P(n) for n=1.

P (1) = 1/21 = 1 – 1/21

or, 1/2 = 1/2

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

Then, 1/2 + 1/4 + 1/8 + … + 1/2k + 1/2k+1

Now, putting the value (i), we get,

= 1 – 1/2k + 1/2k+1   (using eq(i))

= 1 – ((2 – 1)/2k+1)

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

### Question 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)

Step 1:

Now, let us check P(n) for n=1.

P (1) = (2.1 – 1)2 = 1/3 × 1 × (4 – 1)

or, 1 = 1

So, P(1) is true.

Step 2:

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

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

Step 3:

Now, we have to prove for P(k+1) is true.

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

Now, putting the value (i) ,we get,

Or, 1/3 k (4k2 – 1) + (2k + 1)2                         (using eq(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]

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

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

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