# Class 11 NCERT Solutions- Chapter 4 Principal 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 + 3^{2} + …….. + 3^{n-1} =

**Solution:**

We have,

P(n) = 1 + 3 + 3

^{2}+ …….. + 3^{n-1}=For

n=1, we getP(1) = 1 =

So, P(1) is true

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

n=kP(k) = 1 + 3 + 3

^{2}+ …….. + 3^{k-1}= ……………..(1)Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)= 1 + 3 + 3^{2}+ …….. + 3^{k-1}+ 3^{(k+1)-1}= (1 + 3 + 3

^{2}+ …….. + 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

truefor all natural numbers n.

### Question 2: 1 + 2^{3} + 3^{3} + ……….. + n^{3} =

**Solution:**

We have,

P(n) = 1 + 2

^{3}+ 3^{3}+ ……….. + n^{3}=For

n=1, we getP(1) = 1 = = 1

So, P(1) is true

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

n=kP(k) = 1 + 2

^{3}+ 3^{3}+ ……….. + k^{3}= ……………..(1)Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)= 1 + 2^{3}+ 3^{3}+ ……….. + k^{3}+ (k+1)^{3}= (1 + 2

^{3}+ 3^{3}+ ……….. + k^{3}) + (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

truefor all natural numbers n.

### Question 3: 1 + + ……. + =

**Solution:**

We have,

P(n) = 1 + =

For

n=1, we getP(1) = 1 = = 1

So, P(1) is true

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

n=kP(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

truefor 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 getP(1) = 1.2.3 = = = 1.2.3

So, P(1) is true

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

n=kP(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.

### Question 5: 1.3 + 2.3^{2} + 3.3^{3} +…+ n.3^{n} =

**Solution:**

We have,

P(n) = 1.3 + 2.3

^{2}+ 3.3^{3}+…+ n.3^{n}=For

n=1, we getP(1) = 1.3 = 3 = = 3

So, P(1) is true

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

n=kP(k) = 1.3 + 2.3

^{2}+ 3.3^{3}+…+ k.3^{k}= ……………..(1)Let’s prove that P(k + 1) is also true. Now, we have

P(k+1)= 1.3 + 2.3^{2}+ 3.3^{3}+…+ k.3^{k}+ (k+1).3^{(k+1)}= (1.3 + 2.3

^{2}+ 3.3^{3}+…+ 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.truefor 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 getP(1) = 1.2 = 2 = = 2

So, P(1) is true

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

n=kP(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.truefor 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 getP(1) = 1.3 = 3 = = = 3

So, P(1) is true

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

n=kP(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

= + (4k

^{2}+8k+3)=

=

=

=

=

=

=

=

Hence,

P(k+1) =

Thus, P(k + 1) is true, whenever P (k) is true.truefor all natural numbers n.

### Question 8: 1.2 + 2.2^{2} + 3.2^{3} + …+n.2^{n} = (n–1) 2^{n + 1} + 2

**Solution:**

We have,

P(n) = 1.2 + 2.2

^{2}+ 3.2^{3}+ …+n.2^{n}= (n–1) 2^{n + 1}+ 2For

n=1, we getP(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=kP(k) = 1.2 + 2.2

^{2}+ 3.2^{3}+ …+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 + 2.2^{2}+ 3.2^{3}+ …+k.2^{k}+ (k+1).2^{(k+1)}= (1.2 + 2.2

^{2}+ 3.2^{3}+ …+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}+ 2Hence,

P(k+1) = ((k+1)-1).2

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

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

### Question 9: + …… +

**Solution:**

We have,

P(n) =

For

n=1, we getP(1) = = 1 – =

So, P(1) is true

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

n=kP(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.

### Question 10: + …… +

**Solution:**

We have,

P(n) =

For

n=1, we getP(1) =

So, P(1) is true

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

n=kP(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.truefor all natural numbers n.

### Question 11: + …… +

**Solution:**

We have,

P(n) =

For

n=1, we getP(1) =

So, P(1) is true

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

n=kP(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.truefor all natural numbers n.

### Question 12: a + ar + ar^{2} + …… + ar^{n-1} =

**Solution:**

We have,

P(n) = a + ar + ar

^{2}+ …… + 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 + ar

^{2}+ …… + ar^{k-1}= ……………..(1)Let’s prove that P(k + 1) is also true. Now, we have

P(k+1) = a + ar + ar

^{2}+ …… + ar^{k-1}+ ar^{(k+1)-1}= (a + ar + ar

^{2}+ …… + 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.truefor 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 getP(1) = 1+ = 1+3 = 4 = (1+1)

^{2}= 2^{2}= 4So, P(1) is true

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

n=kP(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.truefor all natural numbers n.