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Class 11 NCERT Solutions- Chapter 4 Principal of Mathematical Induction – Exercise 4.1 | Set 2
  • Last Updated : 18 Mar, 2021

Question 14: (1+ \frac{1}{1}) (1+ \frac{1}{2}) (1+ \frac{1}{3}) ..... (1+ \frac{1}{n})   = (n+1)

Solution:

We have,

P(n) = (1+ \frac{1}{1}) (1+ \frac{1}{2}) (1+ \frac{1}{3}) ..... (1+ \frac{1}{n})   = (n+1)

For n=1, we get

P(1) = (1+ \frac{1}{1}  ) = 2 = (1+1) = 2



So, P(1) is true

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

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

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

P(k+1)(1+ \frac{1}{1}) (1+ \frac{1}{2}) (1+ \frac{1}{3}) ..... (1+ \frac{1}{k})   (1+ \frac{1}{k+1})

= ((1+ \frac{1}{1}) (1+ \frac{1}{2}) (1+ \frac{1}{3}) ..... (1+ \frac{1}{k})  (1+ \frac{1}{k+1})

From Eq(1), we get

= (k+1) (1+ \frac{1}{k+1}  )



= (k+1) (\frac{(k+1)+1}{k+1})

= {(k+1)+1}

Hence,

P(k+1) = {(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 15: 12 + 32 + 52 + …… + (2n-1)2\frac{n(2n-1)(2n+1)}{3}

Solution:

We have,

P(n) = 12 + 32 + 52 + …… + (2n-1)2\frac{n(2n-1)(2n+1)}{3}

For n=1, we get



P(1) = 12 = 1 = \frac{1(2(1)-1)(2(1)+1)}{3} = \frac{3}{3}   = 1

So, P(1) is true

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

P(k) = 12 + 32 + 52 + …… + (2k-1)2\frac{k(2k-1)(2k+1)}{3}   ……………..(1)

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

P(k+1) = 12 + 32 + 52 + …… + (2k-1)2 + (2(k+1)-1)2

= (12 + 32 + 52 + …… + (2k-1)2) + (2k+1)2

From Eq(1), we get

\frac{k(2k-1)(2k+1)}{3}   + (2k+1)2

= (2k+1) (\frac{k(2k-1 + 3(2k+1))}{3})



= (2k+1) (\frac{2k^2-k + 6k+3}{3})

= (2k+1) (\frac{2k^2+5k+3}{3})

= (2k+1) (\frac{(2k+3)(k+1)}{3})

(\frac{(k+1)(2k+3)(2k+1)}{3})

(\frac{(k+1) (2(k+1)-1)(2(k+1)+1)}{3})

Hence,

P(k+1) = (\frac{(k+1) (2(k+1)-1)(2(k+1)+1)}{3})

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 16: \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + ..... + \frac{1}{(3n-2)(3n+1)} = \frac{n}{(3n+1)}

Solution:



We have,

P(n) = \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + ..... + \frac{1}{(3n-2)(3n+1)} = \frac{n}{(3n+1)}

For n=1, we get

P(1) = \frac{1}{1.4} = \frac{1}{4} = \frac{1}{(3(1)+1)} = \frac{1}{4}

So, P(1) is true

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

P(k) = \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + ..... + \frac{1}{(3k-2)(3k+1)} = \frac{k}{(3k+1)}   ……………..(1)

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

P(k+1)\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + ..... + \frac{1}{(3k-2)(3k+1)} + \frac{1}{(3(k+1)-2)(3(k+1)+1)}

= (\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + ..... + \frac{1}{(3k-2)(3k+1)}  \frac{1}{(3k+1)(3k+4)}



From Eq(1), we get

\frac{k}{(3k+1)} + \frac{1}{(3k+1)(3k+4)}

\frac{1}{3k+1} (k + \frac{1}{(3k+4)})

\frac{1}{3k+1} (\frac{k(3k+4) +1}{(3k+4)})

\frac{1}{3k+1} (\frac{3k^2+4k+1)}{(3k+4)})

\frac{1}{3k+1} (\frac{(3k+1)(k+1)}{(3k+4)})

\frac{(k+1)}{(3(k+1)+1)}

Hence,

P(k+1) = \frac{(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 17: \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + ..... + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}

Solution:

We have,

P(n) = \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + ..... + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}

For n=1, we get

P(1) = \frac{1}{3.5} = \frac{1}{15} = \frac{1}{3(2(1)+3)} = \frac{1}{15}

So, P(1) is true

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

P(k) = \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + ..... + \frac{1}{(2k+1)(2k+3)} = \frac{k}{3(2k+3)}   ……………..(1)

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



P(k+1)\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + ..... + \frac{1}{(2k+1)(2k+3)} + \frac{1}{(2(k+1)+1)(2(k+1)+3)}

= (\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + ..... + \frac{1}{(2k+1)(2k+3)}  \frac{1}{(2k+3)(2k+5)}

From Eq(1), we get

\frac{k}{3(2k+3)} + \frac{1}{(2k+3)(2k+5)}

\frac{1}{(2k+3)} (\frac{k}{3} + \frac{1}{2k+5})

\frac{1}{(2k+3)} (\frac{k(2k+5)+3)}{3(2k+5)})

\frac{1}{(2k+3)} (\frac{2k^2+5k+3}{3(2k+5)})

\frac{1}{(2k+3)} (\frac{(2k+3)(k+1)}{3(2k+5)})

\frac{k+1}{3(2k+5)}

\frac{k+1}{3(2(k+1)+3)}



Hence,

P(k+1) = \frac{k+1}{3(2(k+1)+3)}

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 18: 1 + 2 + 3 + ….. + n < \frac{1}{8}   (2n+1)2

Solution:

We have,

P(n) = 1 + 2 + 3 + ….. + n < \frac{1}{8}   (2n+1)2

For n=1, we get

P(1) = 1 < \frac{1}{8}   (2(1)+1)2

1 < \frac{9}{8}

So, P(1) is true

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

P(k) = 1 + 2 + 3 + ….. + k < \frac{1}{8}   (2k+1)2 ……………..(1)

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

P(k+1) = 1 + 2 + 3 + ….. + k + (k+1)

= (1 + 2 + 3 + ….. + k) + k+1 < \frac{1}{8}   (2k+1)2 + (k+1)

\frac{1}{8}   ((2k+1)2 + 8(k+1))

\frac{1}{8}   ((4k2+4k+1) + 8k+8))

\frac{1}{8}   (4k2+12k+9)

\frac{1}{8}   (2k+3)2



P(k+1) < \frac{1}{8}   (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 19: n (n + 1) (n + 5) is a multiple of 3.

Solution:

We have,

P (n) = n (n + 1) (n + 5), which is a multiple of 3

For n=1, we get

p(1) = 1 (1 + 1) (1 + 5) = 12, which is a multiple of 3

So, P(1) is true

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

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

P(k) = k (k + 1) (k + 5) = 3m, where m ∈ N ………… (1)

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

P(k+1) = (k + 1) {(k + 1) + 1} {(k + 1) + 5}

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

By multiplying the terms

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

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

Using Eq(1), we get

= 3m + (k + 1) {2 (k + 5) + (k + 2)}

= 3m + (k + 1) {2k + 10 + k + 2}

= 3m + (k + 1) (3k + 12)

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

= 3 {m + (k + 1) (k + 4)}

= 3 × q                             (where q = {m + (k + 1) (k + 4)} is some natural number)

(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3

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 20: 102n – 1 + 1 is divisible by 11.

Solution:

We have,

P (n): 102n – 1 + 1 is divisible by 11

For n=1, we get

P (1) = 102.1 – 1 + 1 = 11, which is divisible by 11

So, P(1) is true

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

102k – 1 + 1 is divisible by 11

102k – 1 + 1 = 11m, where m ∈ N …………… (1)

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

P(k + 1) = 10 2(k + 1) – 1 + 1

= 102k + 2 – 1 + 1

= 102k + 1 + 1

= 102 (102k-1 + 1 – 1) + 1

= 102 (102k-1 + 1) – 102 + 1

Using Eq(1), we get

= 102. 11m – 100 + 1

= 100 × 11m – 99

Taking out the common terms

= 11 (100m – 9)

= 11r, where r = (100m – 9) is some natural number

10 2(k + 1) – 1 + 1 is divisible by 11

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Question 21: x2n – y2n is divisible by x + y.

Solution:

We have,

P (n) = x2n – y2n is divisible by x + y

For n=1, we get

P (1) = x2 × 1 – y2 × 1 = x2 – y2 = (x + y) (x – y), which is divisible by (x + y)

So, P(1) is true

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

x2k – y2k is divisible by x + y

x2k – y2k = m (x + y), where m ∈ N …… (1)

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

x 2(k + 1) – y 2(k + 1)

= x 2k . x2 – y2k . y2

By adding and subtracting y2k we get

= x2 (x2k – y2k + y2k) – y2k. y2

Using Eq(1), we get

= x2 {m (x + y) + y2k} – y2k. y2

= m (x + y) x2 + y2k. x2 – y2k. y2

Taking out the common terms



= m (x + y) x2 + y2k (x2 – y2)

= m (x + y) x2 + y2k (x + y) (x – y)

So we get

= (x + y) {mx2 + y2k (x – y)}, which is a factor of (x + y)

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Question 22: 32n+2 – 8n – 9 is divisible by 8.

Solution:

We have,

P (n) = 32n + 2 – 8n – 9 is divisible by 8

For n=1, we get

P (1) = 32 × 1 + 2 – 8 × 1 – 9 = 64, which is divisible by 8

So, P(1) is true

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

32k + 2 – 8k – 9 is divisible by 8

32k + 2 – 8k – 9 = 8m, where m ∈ N …… (1)

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

3 2(k + 1) + 2 – 8 (k + 1) – 9

= 3 2k + 2 . 32 – 8k – 8 – 9

By adding and subtracting 8k and 9 we get

= 32 (32k + 2 – 8k – 9 + 8k + 9) – 8k – 17

= 32 (32k + 2 – 8k – 9) + 32 (8k + 9) – 8k – 17

Using Eq(1), we get

= 9. 8m + 9 (8k + 9) – 8k – 17

= 9. 8m + 72k + 81 – 8k – 17

= 9. 8m + 64k + 64

By taking out the common terms

= 8 (9m + 8k + 8)

= 8r, where r = (9m + 8k + 8) is a natural number

So 3 2(k + 1) + 2 – 8 (k + 1) – 9 is divisible by 8

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Question 23: 41n – 14n is a multiple of 27.

Solution:

We have,

P (n) = 41n – 14n is a multiple of 27

For n=1, we get

P (1) = 411 – 141 = 27, which is a multiple by 27

So, P(1) is true

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

41k – 14k is a multiple of 27

41k – 14k = 27m, where m ∈ N …… (1)

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

41k + 1 – 14k + 1

= 41k. 41 – 14k. 14

By adding and subtracting 14k we get

= 41 (41k – 14k + 14k) – 14k. 14

= 41 (41k – 14k) + 41. 14k – 14k. 14

Using Eq(1), we get

= 41. 27m + 14k ( 41 – 14)

= 41. 27m + 27. 14k

By taking out the common terms

= 27 (41m – 14k)

= 27r, where r = (41m – 14k) is a natural number

So 41k + 1 – 14k + 1 is a multiple of 27

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Question 24:  (2n + 7) < (n + 3)2.

Solution:

We have,

P(n) = (2n +7) < (n + 3)2

For n=1, we get

2.1 + 7 = 9 < (1 + 3)2 = 16

So, P(1) is true

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

(2k + 7) < (k + 3)2 … (1)

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

{2 (k + 1) + 7} = (2k + 7) + 2

= {2 (k + 1) + 7}

Using Eq(1), we get

(2k + 7) + 2 < (k + 3)2 + 2

2 (k + 1) + 7 < k2 + 6k + 9 + 2

2 (k + 1) + 7 < k2 + 6k + 11

Here,

k2 + 6k + 11 < k2 + 8k + 16

2 (k + 1) + 7 < (k + 4)2

2 (k + 1) + 7 < {(k + 1) + 3}2

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.




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