Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

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Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers

Question 1.
For all n ≥ 1 , prove that
12 + 22 + 32 +……….+ n2 > \(\frac{n^{3}}{3}\)
Answer:
Let p(n): 12 + 22 + 32 + n2
Put n = 1 ⇒ p(1) = 1 > \(\frac{1}{3}\) which is true.
Assuming that true for p(k)
p(k): 12 + 22 + 32 +……….+ k2 > \(\frac{k^{3}}{3}\)
Let p(k + 1): 12 + 22 + 32 +…….+ k2 + (k + 1)2 > \(\frac{k^{3}}{3}\) + (k + 1)2
Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers 1
Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers 2
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 2.
For all n ≥ 1 , prove that 1 + 2 + 3 +…….+ n < \(\frac{1}{8}\)(2n + 1)2
Answer:
Let p(n): 1 + 2 + 3 +…….+ n ,
Put n = 1 ⇒ p(1) = 1 < \(\frac{9}{8}\) which is true.
Assuming that true for p(k)
p(k): 1 + 2 + 3 +…….+ k < \(\frac{1}{8}\)(2k + 1)2
Let p(k +1): 1 + 2 + 3 +……..+ k + (k +1) < \(\frac{1}{8}\) (2k + 1)2 + (k + 1)
Plus One Maths Principle of Mathematical Induction Three Mark Questions and Answers 3
Hence by using the principle of mathematical induction true for all n ∈ N.

Question 3.
For all n ≥ 1, prove that p(n): 23n – 1 is divisible by 7.
Answer:
p(1): 23(1) – 1 = 8 – 1 = 7 divisible by 7, hence true. Assuming that for p(k)
p(k) : 23k – 1 is divisible by 7.
23k – 1 = 7M
P(k + 1): 23(k + 1) – 1 = 23k + 3 – 1
= 23k23 – 1 = 23k × 8 – 1
= 23k × 8 – 8 + 7 = 8(23k – 1) + 7
= 8(7M) + 7
Hence divisible by 7. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 4.
For all n ≥ 1, prove that p(n): n3 + (n + 1)3 + (n + 2)3 is divisible by 9.
Answer:
p(1): 1 + 23 + 33 = 1 + 8 + 27 = 36 divisible by 9, hence true. Assuming that true for p(k)
p(k): k3 + (k + 1)3 + (k + 2)3 is divisible by 9.
⇒ k3 + (k + 1)3 + (k + 2)3 = 9M
p(k +1 ): (k + 1)3 + (k + 2 )3 + (k + 3)3
= (k +1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27
= [(k + 1)3 + (k + 2)3 + k3] + 9[k2 + 3k + 3]
= 9M + 9[k2 + 3k + 3]
Hence p(k + 1)divisible by 9. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers

Question 1.
For all n ≥ 1, prove that
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 4
Answer:
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 5
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 2.
For all n ≥ 1, prove that
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 6
Answer:
Let
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 7
Assuming that true for p(k)
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 8
Let p(k + 1):
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 9
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 3.
For all n ≥ 1 , prove that 1.2.3 + 2.3.4 +………+ n(n + 1)(n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\).
Answer:
Let p(n): 1.2.3 + 2.3.4 +……..+ n(n + 1)(n + 2),
Put n = 1
p(1) = \(\frac{1(1+1)(1+2)(1+3)}{4}\) = 6 which is true.
Assuming that true for p(k)
p(k): 1.2.3 + 2.3.4 +……..+ k(k + 1)(k + 2) = \(\frac{k(k+1)(k+2)(k+3)}{4}\),
Let p(k + 1)
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 10
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 11
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 4.
For all n ≥ 1, prove that
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 12
Answer:
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 13
Plus One Maths Principle of Mathematical Induction Four Mark Questions and Answers 14
Hence by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 5.
For all n ≥ 1 , prove that p(n): n(n + 1 )(n + 5) is divisible by 3.
Answer:
p(1): 1(1 + 1)(1 + 5) = 12divisible by 3, hence true. Assuming that true for p(k)
p(k): k(k + 1)(k + 5) is divisible by 3.
k(k + 1)(k + 5) = 3M
p(k + 1): (k + 1)(k + 2)(k + 6)
= (k + 1)(k2 + 8k + 12)
= (k + 1)(k2 + 5k + 3k +12)
= (k + 1)[k(k + 5) + 3(k + 6)]
= [k(k + 1)(k + 5) + 3(k + 1)(k + 6)]
= [3M + 3(k + 1)(k + 6)]
= 3[M + (k + 1)(k + 6)]
Hence divisible by 3. Therefore by using the principle of mathematical induction true for all n ∈ N.

Plus One Maths Chapter Wise Questions and Answers Chapter 4 Principle of Mathematical Induction

Question 6.
For all n ≥ 1 , prove that p(n): 2.7n + 3.5n – 5 is divisible by 24.
Answer:
p(1): 2.71 + 3.51 – 5 = 14 + 15 – 5 = 24 divisible by 24, hence true. Assuming that true for p(k)
p(k): 2.7k + 3.5k – 5 is divisible by 24.
⇒ 2.7k + 3.5k – 5 = 24M
p(k + 1): 2.7k + 1 + 3.5k + 1 – 5
= 2.7k.7 + 3.5k.5 – 5
= 2.7k.(6 + 1) + 3.5k.(4 + 1) – 5
= 12.7k + 2.7k + 12.5k + 3.5k – 5
= 12(7k + 5k) + (2.7k + 3.5k) – 5
= 12(7k + 5k) + 24M
7k And 5k are odd numbers, therefore (7k + 5k) will be an even and will be divisible by 24, Hence p(k + 1)divisible by 24. Therefore by using the principle of mathematical induction true for all n ∈ N.

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