Class 11 NCERT Solutions- Chapter 4 Principle of Mathematical Induction – Exercise 4.1 | Set 1

Prove the following by using the principle of mathematical induction for all n ∈ N:

Question 1: 1 + 3 + 3² + …….. + 3^n-1 = 

Solution:

We have,

P(n) = 1 + 3 + 3² + …….. + 3^n-1 = 

For n=1, we get

P(1) = 1 = 

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = 1 + 3 + 3² + …….. + 3^k-1 =   ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 1 + 3 + 3² + …….. + 3^k-1 + 3^(k+1)-1

= (1 + 3 + 3² + …….. + 3^k-1) + 3^k

From Eq(1), we get

=  + 3^k

= 

= 

= 

Hence,

P(k+1) = 

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

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

Question 2: 1 + 2³ + 3³ + ……….. + n³ = 

Solution:

We have,

P(n) = 1 + 2³ + 3³ + ……….. + n³ = 

For n=1, we get

P(1) = 1 =  = 1

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = 1 + 2³ + 3³ + ……….. + k³ =   ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 1 + 2³ + 3³ + ……….. + k³ + (k+1)³

= (1 + 2³ + 3³ + ……….. + k³) + (k+1)³

From Eq(1), we get

=  + (k+1)³

=  + (k+1)³

= 

Taking (k+1)², we get 

= 

= 

= 

= 

Hence,

P(k+1) = 

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

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

Question 3: 1 +  + ……. +  = 

Solution:

We have,

P(n) = 1 +  = 

For n=1, we get

P(1) = 1 =  = 1

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = 1 +  =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 1 +  + 

= (1 + ) + 

From Eq(1), we get

= 

As we know that, 

Sum of first natural number,

1 + 2 + 3 + …… + n = 

So, we get

= 

= 

= 

= 

= 

= 

= 

Hence,

P(k+1) = 

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

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

Question 4: 1.2.3 + 2.3.4 +…+ n(n+1) (n+2) = 

Solution:

We have,

P(n) = 1.2.3 + 2.3.4 +…+ n(n+1) (n+2) = 

For n=1, we get

P(1) = 1.2.3 =  =  = 1.2.3

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = 1.2.3 + 2.3.4 +…+ k(k+1) (k+2) =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 1.2.3 + 2.3.4 +…+ k(k+1) (k+2) + (k+1)(k+1+1) (k+1+2)

= (1.2.3 + 2.3.4 +…+ k(k+1) (k+2)) + (k+1)(k+2) (k+3)

From Eq(1), we get

=   + (k+1)(k+2)(k+3)

= (k+1)(k+2) (k+3) ( + 1)

= 

= 

Hence,

P(k+1) = 

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

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

Question 5: 1.3 + 2.3² + 3.3³ +…+ n.3ⁿ = 

Solution:

We have,

P(n) = 1.3 + 2.3² + 3.3³ +…+ n.3ⁿ = 

For n=1, we get

P(1) = 1.3 = 3 =  = 3

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = 1.3 + 2.3² + 3.3³ +…+ k.3^k =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 1.3 + 2.3² + 3.3³ +…+ k.3^k + (k+1).3^(k+1)

= (1.3 + 2.3² + 3.3³ +…+ k.3^k) + (k+1).3^(k+1)

From Eq(1), we get

=  + (k+1).3^(k+1)

= 

= 3^k+1

= 3^k+1

= 3^k+1

= 3^(k+1)+1

Hence,

P(k+1) = 3^(k+1)+1

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

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

Question 6: 1.2 + 2.3 + 3.4 +…+ n.(n+1) = 

Solution:

We have,

P(n) = 1.2 + 2.3 + 3.4 +…+ n.(n+1) = 

For n=1, we get

P(1) = 1.2 = 2 =  = 2

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = 1.2 + 2.3 + 3.4 +…+ k.(k+1) =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 1.2 + 2.3 + 3.4 +…+ k.(k+1) + (k+1)(k+1+1)

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

From Eq(1), we get

=  + (k+1)(k+2)

= 

= (k+1)(k+2) 

= 

= 

Hence,

P(k+1) = 

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

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

Question 7: 1.3 + 3.5 + 5.7 +…+ (2n–1) (2n+1) =

Solution:

We have,

P(n) = 1.3 + 3.5 + 5.7 +…+ (2n–1) (2n+1) = 

For n=1, we get

P(1) = 1.3 = 3 =  =  = 3

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = 1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1) =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 1.3 + 3.5 + 5.7 +…+ (2k–1) (2k+1) + (2(k+1)-1)(2(k+1)+1)

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

From Eq(1), we get

=  + (4k²+8k+3)

= 

= 

= 

=

= 

= 

= 

= 

Hence,

P(k+1) = 

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

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

Question 8: 1.2 + 2.2² + 3.2³ + …+n.2ⁿ = (n–1) 2^{n + 1} + 2

Solution:

We have,

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

For n=1, we get

P(1) = 1.2 = 2 = (1–1) 2(1) + 1 + 2 = 2

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

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

Let’s prove that P(k + 1) is also true. Now, we have

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

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

From Eq(1), we get

= (k–1) 2^{k + 1} + 2 + (k+1).2^k+1

= 2^{k + 1}((k–1) + (k+1)) + 2 

= 2k + 1(2k) + 2 

= k.2^k+1+1 + 2 

= ((k+1)-1).2^(k+1)+1 + 2 

Hence,

P(k+1) = ((k+1)-1).2^(k+1)+1 + 2 

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

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

Question 9:  + …… + 

Solution:

We have,

P(n) = 

For n=1, we get

P(1) =  = 1 –  = 

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 

= () + 

From Eq(1), we get

= 1 – 

= 1 – 

= 1 – 

= 1 – 

= 1 –

Hence,

P(k+1) = 1 – 

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

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

Question 10:  + …… + 

Solution:

We have,

P(n) = 

For n=1, we get

P(1) = 

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 

= () + 

From Eq(1), we get

= 

= 

= 

= 

= 

= 

= 

Hence,

P(k+1) = 

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

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

Question 11:  + …… + 

Solution:

We have,

P(n) = 

For n=1, we get

P(1) = 

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 

= () + 

From Eq(1), we get

= 

= 

= 

= 

= 

= 

= 

= 

= 

= 

Hence,

P(k+1) = 

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

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

Question 12: a + ar + ar² + …… + ar^n-1 = 

Solution:

We have,

P(n) = a + ar + ar² + …… + ar^n-1 = 

For n=1, we get

P(1) = a =  = a

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) = a + ar + ar² + …… + ar^k-1 =  ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = a + ar + ar² + …… + ar^k-1 + ar^(k+1)-1

= (a + ar + ar² + …… + ar^k-1) + ar^k

From Eq(1), we get

=  + ar^k

= 

= 

= 

Hence,

P(k+1) = 

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

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

Question 13: (1+ ) (1+ ) (1+ ) ….. (1+ ) = (n+1)²

Solution:

We have,

P(n) =  = (n+1)²

For n=1, we get

P(1) = 1+  = 1+3 = 4 = (1+1)² = 2² = 4

So, P(1) is true

Assume that P(k) is true for some positive integer n=k

P(k) =  = (k+1)² ……………..(1)

Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = 

= 

From Eq(1), we get

= (k+1)² (1+)

= (k+1)²

= (k+1)² + 2(k+1) + 1

= {(k+1)}²

Hence,

P(k+1) = {(k+1)}²

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

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