Question 1: Let P (n) be the statement 7 divides (2^{3n} -1) . What is P (n + 1) ?

Answer:

P (n + 1) is the statement 7 divides (2^{3(n+ 1)} - 1) .

Clearly, P (n + 1) is obtained by replacing n by (n+ 1) in P (n) .

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Question 2: If P (n) is the statement n^2 > 100 , prove that whenever P(r) is true, P (r +1) is also true.

Answer:

The statement P (n) is n^2 >100 . Let P (r) be true. Then r^2 >100

We wish to prove that the statement P (r +1) is true i.e. (r +1)^2 >100

Now,

P (r) is true

\Rightarrow r^2 > 100

\Rightarrow r^2 + 2r + 1> 100 + 2r + 1 [ adding (2r+1) on both sides]

\Rightarrow (r+1)^2 > 100 + 2r + 1

\Rightarrow (r+1)^2 > 100

\Rightarrow P(r+1) is true [ \because 100 + 2r + 1 > 100 for every natural number r ]

Thus, whenever P(r) is true, P (r + 1) is also true.

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Question 3: Let P(n) be the statement 3^n > n . If P(n) is true, prove that P(n+1) is true.

Answer:

We are given that P (n) is true i.e. 3^n >n and we wish to prove that P (n + 1 ) is true i.e. 3^(n+ 1) >(n+1) .

Now,

P (n) is true

\Rightarrow 3^n > n

\Rightarrow 3.3^n > 3n [Multiplying both sides by 3 ]

\Rightarrow 3^{n+1} > n + 2n

\Rightarrow 3^{n+1} > n + 1 [ \because 2n> 1 for every n \in N \Rightarrow 2n+n>n+1 . for every n \in N ]

\Rightarrow P (n+1) is true

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Question 4: lf P (n) is the statement 2^{3n} - 1 is an integral multiple of 7 , and if P(r) is true, prove that P (r + 1) is true.

Answer:

Let P (r) be true. Then, 2^{3r} -1 is an integral multiple of 7 .

We wish to prove that P (r + 1) is true i.e. 2^{3(r+ 1)} - 1 is an integral multiple of 7 .

Now,

P(r) is true

\Rightarrow 2^{3r}-1 . is an integral multiple of 7

\Rightarrow 2^{3r}-1 = 7 \lambda , for some \lambda \in N

\Rightarrow 2^{3r} = 7 \lambda + 1

Now, 2^{3(r + 1)}-1 = 2^{3r} \times 2^3 -1, = (7 \lambda +1) \times 8-1

\Rightarrow 2^{3(r+1)} -1 = 56 \lambda +8-1 = 56 \lambda +7 =7 (8\lambda+1)

\Rightarrow 2^{3(r+1)}-1 =7 \mu , where \mu =8 \lambda +1 \in N

\Rightarrow 2^{3(r + l)}-1 is an integral multiple of 7

\Rightarrow P(r+1) is true

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Question 5: Prove by the principle of mathematical induction that for all n \in N, n^2 + n is even natural number.

Answer:

Let P (n) be the statement n^2 + n is even

We have, P (n): n^2 + n is even

\because 1^2 +1 = 2 , which is even

\therefore P (1) is true

Let P (m) be true. Then,

P(m) is ture \Rightarrow ,m^2 + m is even \Rightarrow m^2 +m = 2\lambda for some \lambda \in N

Now, we shall show thal P (m+ 1) is true. For this we have to show that (m + 1)^2 + (m+1) is an even natural number.

\Rightarrow (m+1)^2 +(m+1) = (m^2 +2m + 1) + (m+1) = (m^2 + m )+(2m+2)

\Rightarrow (m+1)^2 +(m+1) = m^2 + m + 2(m+1) = 2\lambda +2(m+1)

\Rightarrow (m+1)^2 +(m+1) = 2( \lambda +m+1) = 2\mu ,where \mu = \lambda +m+1 \in N

\Rightarrow (m + 1)^2 + (m + 1) is an even natural number

\Rightarrow P(m+1) is true

Thus, P (m) is true \Rightarrow P (m + 1) is true

Hence,by the principle of mathematical induction, P (n) is true for all n \in N i.e. n^2 +n is even for all n \in N .

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Question 6: Prove by the principle of mathematical induction that n (n + 1) (2n + 1) is divisible by 6 for all n \in N

Answer:

Let P (n) be the statement n (n + 1) (2n + 1) is divisible by 6

i.e. P (n) : n (n + 1) (2n + 1) is divisible by 6

We have, P (1) : 1 (1 + 1) (2 + 1) is divisible by 6

\because 1 (1 + 1) (2 + 1) = 6 which is divisibleby 6

\therefore P(1 ) is true

Let P(m) be true. Then,

m (m + 1) (2m + 1) is divisible by 6

m (m + 1) (2m + 1) = 6 \lambda , for some \lambda \in N

Now, we shall show that P (m + 1) is true. For this we have to show that (m+1) (m + 1 + 1) [ 2(m + 1) + 1 ] is divisibleby 6

(m+1)(m+ 1 + 1) [ 2(m+1) + 1 ] = (m+1)(m+2) [(2m+ 1) + 2 ]

= (m + 1) (m + 2) (2m + 1) + 2 (m + 1) (m + 2)

= m (m + 1) (2m + 1) + 2 (m + 1) (2m + 1) + 2 (m + 1) (m + 2)

= m (m + 1) (2m + 1) + 2 (m + 1) (2m + 1 + m + 2)

= m (m + 1) (2m + 1) + 2 (m + 1) (3m + 3)

= m(m+1)(2m+1)+6 (m+1)^2 = 6\lambda + 6(m+1)^2

= 6 { \lambda + ( m+1)^2 } which is divisible by 5

\Rightarrow P (m + 1) is true

Thus, P (m) is true \Rightarrow P (m + 1) is true

Hence, by the principle of mathematical induction, the given statement is true for all n \in N

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Question 7: Prove by induction that the sum S_n =n^3 + 3n^2 + 5n + 3 is divisible by 3 for all n \in N

Answer:

Let P (n) be the statement given by P(n): S_n =n^3 + 3n^2 + 5n + 3 is divisible by 3

Wehave, P(1):S_1 = 1^3 + 3(1)^2 +5(1) + 3 = 12 is divisible by 3

\therefore P (1) is true

Let P(m) be true. Then,

S_n =m^3 + 3m^2 + 5m + 3 is divisible by 3

S_n =m^3 + 3m^2 + 5m + 3 = 3 \lambda for \lambda \in N

We now wish to show that P (m+ 1) is true. For this we have to show that (m +1)^3 + 3 (m+ 1)^2 + 5 (m +1) + 3 is divisible by 3

Now, (m +1)^3 + 3 (m + 1)^2 + 5 (m+ 1) + 3

= (m^3 + 3 m^2 + 5 m + 3) + 3m^2 + 9m + 9

= 3 \lambda + 3(m^2 + 3m+ 3)

= 3( \lambda + m^2+3m+3)

= 3 \mu where \mu = \lambda + m^2+3m+3 \in N

\therefore P (m + 1) is true

Thus, P (m) is true \Rightarrow  P (m + 1) is true

Hence, by the principle of mathematical induction the statement is true for all n \in N

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Question 8: Using principle of mathematical induction, prove that x^{2n} - y^{2n} it divisible by x + y for all n \in N

Answer:

Let P (n) be the statement given by P(n) : x^{2n} - y^{2n} is divisible by (x + y)

P (1) = (x^2 - y^2) = (x-y)(x+y) which is divisible by (x + y)

So, P(1) is true.

Let P (m) be true. Then,

x^{2m} - y^{2m} is divisible by x + y

\Rightarrow x^{2m} - y^{2m} = \lambda ( x+y)

We shall now show that P (m+ 1) is true i.e. x^{2m+2} - y^{2m+2} is divisible by (x + y)

Now x^{2m+2} - y^{2m+2} = x^{2m+2} - x^{2m}y^2 + x^{2m}y^2 - y^{2m+2}

= x^{2m} (x^2 - y^2) + y^2 ( x^{2m} - y^{2m})

= x^{2m} (x^2 - y^2) + y^2 \lambda (x+y)

= (x+y) [ x^{2m} (x-y) + \lambda y^2 ]

Clearly, it is divisible by (x + y)

\therefore P (m + 1) is true

Thus, P (m) is true \Rightarrow P (m + 1) is true.

Hence , by the principle of mathematical induction P (n) is true for all n \in N

i.e. x^{2n} - y^{2n} it divisible by (x + y) for all n \in N

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Question 9: Using the principle of mathematical induction, prove that (2^{3n}-1) is divisible by 7 for all n \in N

Answer:

Let P (n) be the statement given by P (n) : 2^{3n} -1 is divisible by 7

P (1) :2^{3 \times 1} - 1 is divisible by 7

Clearly, 2^{3 \times 1} -1= 8 - 1 = 7 which is divisibte by 7

So, P (1) is true

Let P (m) be true. Then,

2^{3m} -1 is divisible by 7

\Rightarrow 2^{3m} -1 = 7 \lambda for \lambda \in N

We shall now show that P (m+ 1) is true. For this we have to show that 2^{3(m + 1)} -1 is divisible by 7

Now, 2^{3(m+ 1)} -1 = 2^{3m} \times 2^3 -1 = (7 \lambda + 1) 2^3 -1 = 56 \lambda + 8 - 1 = 7 (8 \lambda + 1) is divisible by 7

\therefore P (m+ 1) is true

\therefore P (m) is true = P (m + 1) is true

Hence , by the principle of mathematical induction, P(n) is true for all n \in N \Rightarrow (2^{3n}-l) is divisible by 7 for all n \in N

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Question 10: Prove that: 2.7^n + 3.5^n - 5 is divisible by 24 , for all n \in N

Answer:

Let P (n) be the statement given by P(n) : 2.7^n + 3.5^n - 5 is divisible by 24 , for all n \in N

We have, P(1) = 2.7^1 + 3.5^1 - 5 = 14 + 15 - 5 = 24 which is divisible by 24

\therefore P (1) is true

Let P (m) be true. Then, 2.7^m + 3.5^m - 5 is divisible by 24 , for all m \in N

\Rightarrow 2.7^m + 3.5^m - 5 = 24 \lambda for some \lambda \in N

\Rightarrow 3.5^m = 24 \lambda + 5 - 2.7^m

We shall now show that P (m+ 1) is true. For this we have to show that 2.7^{m+1} + 3.5^{m+1} - 5 is divisible by 24

2.7^{m+1} + 3.5^{m+1} - 5

= 2.7^{m+1} + (3.5^m)5 - 5

= 2.7^{m+1} + (24 \lambda + 5 - 2.7^m)5 - 5

= 2.7^{m+1} + 120 \lambda + 25 - 10. 7^m - 5

= ( 2.7^{m+1} - 10. 7^m) + 120 \lambda + 20

= ( 2.7^{m+1} - 10. 7^m) + 120 \lambda + 24-4

= ( 14 - 10) 7^m - 4 + 24 ( 5 \lambda + 1)

= 4 ( 7^m - 1) + 24 ( 5 \lambda + 1)

= 4 \times 6\mu + 24 ( 5 \lambda + 1)

[ Since 7^m - 1 is a multiple of 6 for all m \in N \therefore 7^m - 1 = 6 \mu for all \mu \in N ]

= 24 ( \mu + 5 \lambda + 1 ) which is divisible by 24

Therefore P (m + 1) is true

Thus, P (m) is true \Rightarrow P (m + 1) is true

Hence, by the principle of mathematical induction, P (n) is true for all n \in N

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Question 11: Prove by induction that (2n+7)<(n+3)^2 for all natural numbers n . Using this, prove by induction that (n+ 3)^2 \leq 2^{n+3} for all n \in N

Answer:

Let P (n) be the statement given by P(n) : (2n+7)<(n+3)^2

P(1) = ( 2 \times 1 + 7) < ( 1 + 3 )^2 \Rightarrow 9 < 16

Therefore P (1) is true

Let P (m) be true. Then, (2m+7)<(m+3)^2 for all natural numbers m

We shall now show that P (m + 1) is true whenever P (m) is true. For this we have to show that (2(m+1)+7)<((m+1)+3)^2

Now, P (m) is true

(2m+7)<(m+3)^2

\Rightarrow (2m+7) + 2 <(m+3)^2 + 2

\Rightarrow 2(m+1) +7 < m^2 + 6m+11

\Rightarrow 2(m+1)+7 <m^2 + 6m+11 < m^2 +8m+16

\Rightarrow 2(m+1)+7 <(m+4)^2

\Rightarrow [2(m+1) +7] < [(m+1) + 3]^2

\Rightarrow P (m + 1) is true

Hence, by the principle of mathematical induction P (n) is true for all n \in N

Let P'(n) be the statement given by P'(n) :(n+ 3)^2 \leq 2^{n+3} for all n \in N

P'(1) : (1 + 3)^2 \leq 2^{1+ 3}

\because (1 + 3)^2 = 16 \leq 2^{1+ 3}

\therefore P'(1) is true

Let P'(m) be true. Then, (m+ 3)^2 \leq 2^{m+3} for all natural numbers m

We shall now show that P' (m + 1) is true whenever P' (m) is true. For this we have to show that ((m+1)+3)^2 \leq 2^{(m+1)+3}

Now, P' (m) is true

\Rightarrow (m+ 3)^2 \leq 2^{m+3}

\Rightarrow (m+ 3)^2 + ( 2m + 7) \leq 2^{m+3} + ( 2m + 7)

\Rightarrow (m+4)^2 \leq 2^{m+3} + ( m+3)^2

\Rightarrow (m+4)^2 \leq 2^{m+3} + 2^{m+3}

\Rightarrow (m+4)^2 \leq 2.2^{m+3}

\Rightarrow (m+4)^2 \leq 2^{m+4}

\Rightarrow [ (m+1)+3]^2 \leq 2^{(m+1)+3}

\Rightarrow P'(m+1) is true

Hence by the principle of mathematical induction P'(n) is true for all n \in N i.e. (n+ 3)^2 \leq 2^{n+3} for all n \in N