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

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

Solution:

We have,

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

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

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

= (1 + 3 + 32 + …….. + 3k-1) + 3k

From Eq(1), we get

+ 3k

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 +â€¦+ n(n+1) (n+2) =

Solution:

We have,

P(n) = 1.2.3 + 2.3.4 +â€¦+ 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 +â€¦+ 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 +â€¦+ k(k+1) (k+2) + (k+1)(k+1+1) (k+1+2)

= (1.2.3 + 2.3.4 +â€¦+ 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 +â€¦+ n.3n =

Solution:

We have,

P(n) = 1.3 + 2.32 + 3.33 +â€¦+ 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 +â€¦+ k.3k ……………..(1)

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

P(k+1) = 1.3 + 2.32 + 3.33 +â€¦+ k.3k + (k+1).3(k+1)

= (1.3 + 2.32 + 3.33 +â€¦+ k.3k) + (k+1).3(k+1)

From Eq(1), we get

+ (k+1).3(k+1)

= 3k+1

= 3k+1

= 3k+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 +â€¦+ n.(n+1) =

Solution:

We have,

P(n) = 1.2 + 2.3 + 3.4 +â€¦+ 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 +â€¦+ 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 +â€¦+ k.(k+1) + (k+1)(k+1+1)

= (1.2 + 2.3 + 3.4 +â€¦+ 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 +â€¦+ (2nâ€“1) (2n+1) =

Solution:

We have,

P(n) = 1.3 + 3.5 + 5.7 +â€¦+ (2nâ€“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 +â€¦+ (2kâ€“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 +â€¦+ (2kâ€“1) (2k+1) + (2(k+1)-1)(2(k+1)+1)

= (1.3 + 3.5 + 5.7 +â€¦+ (2kâ€“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 = (nâ€“1) 2n + 1 + 2

Solution:

We have,

P(n) = 1.2 + 2.22 + 3.23 + …+n.2n = (nâ€“1) 2n + 1 + 2

For n=1, we get

P(1) = 1.2 = 2 = (1â€“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.2k = (kâ€“1) 2k + 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.2k + (k+1).2(k+1)

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

From Eq(1), we get

= (kâ€“1) 2k + 1 + 2 + (k+1).2k+1

= 2k + 1((kâ€“1) + (k+1)) + 2

= 2k + 1(2k) + 2

= k.2k+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 + …… + arn-1 =

Solution:

We have,

P(n) = a + ar + ar2 + …… + arn-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 + …… + ark-1 ……………..(1)

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

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

= (a + ar + ar2 + …… + ark-1) + ark

From Eq(1), we get

+ ark

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.

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