Mathematical Induction Principles Questions and Answers

This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Principle of Mathematical Induction”.

1. What is the base case for the inequality 7ⁿ > n³, where n = 3?
a) 652 > 189
b) 42 < 132
c) 343 > 27
d) 42 <= 431
View Answer

Answer: c
Explanation: By the principle of mathematical induction, we have 7³ > 3³ ⇒ 343 > 27 as a base case and it is true for n = 3.

2. In the principle of mathematical induction, which of the following steps is mandatory?
a) induction hypothesis
b) inductive reference
c) induction set assumption
d) minimal set representation
View Answer

Answer: a
Explanation: The hypothesis of Step is a must for mathematical induction that is the statement is true for n = k, where n and k are any natural numbers, which is also called induction assumption or induction hypothesis.

3. For m = 1, 2, …, 4m+2 is a multiple of ________
a) 3
b) 5
c) 6
d) 2
View Answer

Answer: d
Explanation: For n = 1, 4 * 1 + 2 = 6, which is a multiple of 2. Assume that 4m+2 is true for m=k and so 4k+2 is true based on the assumption. Now, to prove that 4k+2 is also a multiple of 2 ⇒ 4(k+1)+2 ⇒ 2 * 4k – 4k + 6 ⇒ 2*4k+4 – 4k+2 ⇒ 2(4k+2) – 2(2k+1). Here, the first term 2(4k+2) is true as per assumption and the second term 2(4k+2) is must to be a multiple of 2. Hence, 4(k+1)+2 is a multiple of 2. So, by induction hypothesis, (4m+2) is a multiple of 2, for m = 1,2,3,…

4. For any integer m>=3, the series 2+4+6+…+(4m) can be equivalent to ________
a) m²+3
b) m+1
c) m^m
d) 3m²+4
View Answer

Answer: a
Explanation: The required answer is m²+3. Now, by induction assumption, we have to prove 2+4+6+…+4(k+1) = (k+1)²+3 also can be true, 2+4+6+…+4(k+1) = 2+4+6+⋯+(4k+4) and by the subsequent steps, we can prove that (m+1)²+3 also holds for m=k. So, it is proved.

5. For every natural number k, which of the following is true?
a) (mn)^k = m^kn^k
b) m*k = n + 1
c) (m+n)^k = k + 1
d) m^kn = mn^k
View Answer

Answer: a
Explanation: In the first step, for k = 1, (mn)¹ = m¹n¹ = mn, hence it is true. Let us assume the statement is true for k = l, Now by induction assumption, (mn)¹ = m¹n¹ is true. So, to prove, (mn)^l+1 = m^l + 1n^l+1, we have (mn)^l = m^ln^l and multiplying both sides by (mn) ⇒ (mn)¹(mn)=(m¹n¹)(mn)
⇒ (mn)^l+1 = (mm¹)(nn¹) ⇒ (mn)^l+1 = (m^l+1n^l+1). Hence, it is proved. So, (mn)^k = m^kn^k is true for every natural number k.

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6. By induction hypothesis, the series 1² + 2² + 3² + … + p² can be proved equivalent to ____________
a) \(\frac{p^2+2}{7}\)
b) \(\frac{p*(p + 1)*(2p + 1)}{6}\)
c) \(\frac{p*(p+1)}{4}\)
d) p+p²
View Answer

Answer: b
Explanation: By principle of mathematical induction, we now assume that p (b) is true 1² + 2² + 3² + … + b² = \(\frac{b (b + 1) (2b + 1)}{6}\)
so to prove P(b+1): 1² + 2² + 3² + … + b² + (b + 1)² = \(\frac{b (b + 1) (2b + 1)}{6}\) + (b + 1)²
By induction assumption it is shown that 1² + 2² + 3² + … + b² + (b + 1)² = \(\frac{(b + 1) [(b + 2) (2b + 3)]}{6}\). Hence it is proved that 1² + 2² + 3² + … + p² = \(\frac{p*(p + 1)*(2p + 1)}{6}\).

7. For any positive integer m ______ is divisible by 4.
a) 5m² + 2
b) 3m + 1
c) m² + 3
d) m³ + 3m
View Answer

Answer: d
Explanation: The required answer is, m³ + 3m. Now, by induction hypothesis, we have to prove for m=k, k³+3k is divisible by 4. So, (k + 1)³ + 3 (k + 1) = k³ + 3 k² + 6 k + 4
= [k³ + 3 k] + [3 k² + 3 k + 4] = 4M + (12k² + 12k) – (8k² + 8k – 4), both the terms are divisible by 4. Hence (k + 1)³ + 3 (k + 1) is also divisible by 4 and hence it is proved for any integer m.

8. According to principle of mathematical induction, if P(k+1) = m^(k+1) + 5 is true then _____ must be true.
a) P(k) = 3m^(k)
b) P(k) = m^(k) + 5
c) P(k) = m^(k+2) + 5
d) P(k) = m^(k)
View Answer

Answer: b
Explanation: By the principle of mathematical induction, if a statement is true for any number m = k, then for its successor m = k + 1, the statement also satisfies, provided the statement is true for m = 1. So, the required answer is p(k) = m^k + 5.

9. Which of the following is the base case for 4ⁿ⁺¹ > (n+1)² where n = 2?
a) 64 > 9
b) 16 > 2
c) 27 < 91
d) 54 > 8
View Answer

Answer: a
Explanation: Statement By principle of mathematical induction, for n=2 the base case of the inequation 4ⁿ⁺¹ > (n+1)² should be 64 > 9 and it is true.

10. What is the induction hypothesis assumption for the inequality m ! > 2^m where m>=4?
a) for m=k, k+1!>2^k holds
b) for m=k, k!>2^k holds
c) for m=k, k!>3^k holds
d) for m=k, k!>2^k+1 holds
View Answer

Answer: b
Explanation: By the induction hypothesis, assume that p (k) = k! > 2^k is true, for m=k and we need to prove this by the principle of mathematical induction.

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