Principle of Mathematical Induction — Class 11 MCQs with Answers
Class 11 CBSE Mathematics · Chapter 4
66 practice questions · 22 Easy · 22 Medium · 22 Hard · Updated
Practise the most important Class 11 CBSE Mathematics questions from Chapter 4, "Principle of Mathematical Induction". You get 6 timed quizzes made from 66 NCERT-based MCQs, with answers and explanations. The questions are split into 22 Easy, 22 Medium and 22 Hard. Warm up on the basics, then move on to the exam-level questions that set top scorers in CBSE & Maharashtra HSC Board exams, JEE Main, MHT-CET and JEE Advanced apart.
To score well in "Principle of Mathematical Induction", focus on fast problem-solving, formula recall and step-by-step working. Each MCQ here is timed and uses exam-style marking (+4 correct, −1 wrong, 0 skipped). This trains you to stay accurate under time pressure, as real papers need. Every question has a short explanation, so a wrong answer becomes a quick lesson. It is the fastest way to fix gaps before a test.
Use this chapter for focused revision. Start with the Easy set to check your basics on Principle of Mathematical Induction, then move to Medium and Hard to practise applying them. Your accuracy, streaks and XP save automatically. This chapter also adds to your overall Class 11 Mathematics mastery score. 13 sample questions are solved in full below, with the answer and a worked explanation. Sign in free to start practising.
Key concepts: Principle of Mathematical Induction (Class 11 Mathematics)
Mathematical induction proves that a statement P(n) holds for every natural number by a two-step domino argument: verify a base case, then show each case forces its successor. This chapter formalises the principle and applies it to sums of series, divisibility and inequalities, stressing that both steps are indispensable.
- The principle
- If P(1) is true and P(k) true always implies P(k+1) true, then P(n) is true for every natural number n. Two conditions yield one conclusion.
- Base step
- Verify the statement for the smallest value, usually n=1. Without a true starting domino, the whole chain of implications proves nothing.
- Inductive hypothesis
- The assumption that P(k) is true for some arbitrary fixed natural number k. It is assumed, not proved, and becomes the tool for the next step.
- Inductive step
- Using the hypothesis P(k), prove P(k+1) by algebra. The whole argument hinges on genuinely using the assumed case, not re-deriving it.
Principle of Mathematical Induction — important questions & MCQs with answers (Class 11 Mathematics)
13 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 53 are timed and scored when you sign in.
- Q1Easy
First step in PMI is:
A.ConcludeB.Assume P(k)C.Prove P(k+1)D.Verify base case (n=1)✓ CorrectAnswer: D. Verify base case (n=1)
Explanation: PMI's first step is the base case: verify P(1) holds directly, before assuming anything. Jumping straight to ‘assume P(k)’ skips this and starts at the inductive step instead, which comes second.
- Q2Easy
Sum of first n natural numbers:
A.n(n−1)/2B.n²C.n(n+1)/2✓ CorrectD.n+1Answer: C. n(n+1)/2
Explanation: Sum 1+2+...+n = n(n+1)/2 — check n=4: 1+2+3+4=10, and 4×5/2=10 too. Confusing it with n(n−1)/2, which shifts every term down by 1, undercounts the true sum.
- Q3Easy
Principle of mathematical induction is used to prove statements about:
A.Natural numbers✓ CorrectB.Real numbersC.IntegersD.Rationals onlyAnswer: A. Natural numbers
Explanation: PMI proves statements P(n) indexed by natural numbers n=1,2,3,..., since the base-then-step logic needs a discrete starting point and successor. It does not apply directly to continuous sets like the reals.
- Q4Easy
Sum 1 + 2 + 3 + ... + n =
A.n²B.n(n−1)/2C.n(n+1)/2✓ CorrectD.2nAnswer: C. n(n+1)/2
Explanation: Sum 1+2+...+n = n(n+1)/2 — check n=5: 1+2+3+4+5=15, and 5×6/2=15. This formula, proved by induction, underlies the sum-of-squares and sum-of-cubes identities too.
- Q5Easy
Step 1 of induction is to verify P(n) for:
A.Only n = 0B.All nC.No nD.The base case (usually n = 1)✓ CorrectAnswer: D. The base case (usually n = 1)
Explanation: Step 1 of induction verifies the base case — usually P(1), though for some statements the smallest true n is 4 or 5 instead. Without this check, the inductive chain has no starting link.
- Q6Medium
Inductive step requires assuming truth for:
A.n=1 onlyB.All nC.n=k, then proving for n=k+1✓ CorrectD.n=k+1 onlyAnswer: C. n=k, then proving for n=k+1
Explanation: The inductive step assumes P(k) is true for some k, then proves P(k+1) follows from it. Assuming P(k+1) instead is circular — that would mean assuming exactly what needs proving.
- Q7Medium
1²+2²+...+n² =
A.n²B.n(n+1)(2n+1)/6✓ CorrectC.(n(n+1)/2)²D.n(n+1)Answer: B. n(n+1)(2n+1)/6
Explanation: Σk² from 1 to n is n(n+1)(2n+1)/6 — check n=3: 1+4+9=14, and 3×4×7/6=14. Mixing this with the sum-of-cubes formula [n(n+1)/2]² is the usual mix-up.
- Q8Medium
Sum 1 + 2 + ... + 100 equals:
A.10000B.100C.5050✓ CorrectD.505Answer: C. 5050
Explanation: Sum 1+2+...+100 = n(n+1)/2 with n=100: 100×101/2 = 10100/2 = 5050. Mixing up the division step, or using n² instead, gives wrong distractors like 10000.
- Q9Medium
By induction, 3 divides (5ⁿ − 2ⁿ) for all n ≥ 1. Base case n = 1:
A.Cannot verifyB.3 = 0C.5 − 2 = 3, divisible✓ CorrectD.Always falseAnswer: C. 5 − 2 = 3, divisible
Explanation: For 3 | (5ⁿ−2ⁿ), the base case n=1 gives 5−2=3, which is divisible by 3 — confirming the base before the inductive step assumes P(k) and derives P(k+1) from it.
- Q10Medium
Sum 1² + 2² + ... + 10² equals:
A.550B.100C.45D.385✓ CorrectAnswer: D. 385
Explanation: Σk² from 1 to 10 = n(n+1)(2n+1)/6 = 10×11×21/6 = 2310/6 = 385. Using the plain sum formula n(n+1)/2 = 55 instead would give the wrong distractor 55, not the squared sum.
- Q11Hard
n(n²+5) is divisible by:
A.6✓ CorrectB.5C.10D.7Answer: A. 6
Explanation: n(n²+5) = n(n−1)(n+1) + 6n: the first term is 3 consecutive integers (divisible by 6), the second a multiple of 6, so the whole expression is always divisible by 6, not just 3 or 5.
- Q12Hard
By induction, n³ + 2n is divisible by:
A.2B.3✓ CorrectC.6D.9Answer: B. 3
Explanation: n³+2n = n(n−1)(n+1) + 3n: the first part is 3 consecutive integers (always divisible by 3), the second a multiple of 3, so the whole sum is always divisible by 3 — check n=3: 27+6=33=3×11.
- Q13Hard
Prove 2ⁿ ≥ n + 1 for n ≥ 1. Base n = 1: 2 ≥ 2 ✓. Inductive: assume 2ᵏ ≥ k + 1. Show 2ᵏ⁺¹ ≥ k + 2.
A.Inductive step failsB.Cannot proveC.2ᵏ⁺¹ = 2ᵏ + 1D.2ᵏ⁺¹ = 2·2ᵏ ≥ 2(k+1) ≥ k+2 for k ≥ 0✓ CorrectAnswer: D. 2ᵏ⁺¹ = 2·2ᵏ ≥ 2(k+1) ≥ k+2 for k ≥ 0
Explanation: Assuming 2ᵏ≥k+1, doubling both sides gives 2ᵏ⁺¹≥2(k+1)=2k+2. Since 2k+2≥k+2 whenever k≥0, the chain 2ᵏ⁺¹≥2k+2≥k+2 completes the inductive step.
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What are the key concepts in Class 11 Mathematics Principle of Mathematical Induction?+
Mathematical induction proves that a statement P(n) holds for every natural number by a two-step domino argument: verify a base case, then show each case forces its successor. This chapter formalises the principle and applies it to sums of series, divisibility and inequalities, stressing that both steps are indispensable. Key ideas include The principle, Base step, Inductive hypothesis, Inductive step.
What does Class 11 Mathematics Chapter 4 (Principle of Mathematical Induction) cover on XamBaaz?+
It has 66 NCERT-based MCQs on "Principle of Mathematical Induction": 22 Easy, 22 Medium and 22 Hard. Together they make 6 timed quizzes, and you never get the same set twice. Every question has an instant explanation. They help you prepare for CBSE & Maharashtra HSC Board exams, JEE Main, MHT-CET and JEE Advanced.
Are these "Principle of Mathematical Induction" questions free to practise?+
Yes. Sign in with Google to practise "Principle of Mathematical Induction" free. Full unlimited access is ₹999/year on a launch offer until 1 December 2026. No chapter is charged separately.
How should I revise "Principle of Mathematical Induction" for the exam?+
Start with the Easy quiz to check your basics, then try Medium and Hard to practise applying them. There are 6 timed quizzes on this chapter, so you can come back for a fresh set instead of one you have seen. Read each explanation, retry the questions you miss, and track your accuracy until it stays high.
Are these "Principle of Mathematical Induction" MCQs available with answers?+
Yes. 13 sample questions are shown here in full, each with the correct option and a step-by-step "Why" explanation. Sign in free with Google to start practising, with instant scoring.
Is there negative marking in the "Principle of Mathematical Induction" quizzes?+
Yes. The timed quizzes use exam-style marking: +4 for a right answer, −1 for a wrong one and 0 for a skip, the same negative marking as JEE Main and JEE Advanced. MHT-CET and CBSE board papers have no negative marking. Our mocks for those are scored their way.
What are the important questions from Principle of Mathematical Induction (Class 11 Mathematics)?+
The questions that matter most test The principle, Base step, Inductive hypothesis, Inductive step. This page shows 13 solved important MCQs with answers and explanations; all 66 questions on the chapter are available as timed quizzes once you sign in.
Is there an online quiz for Principle of Mathematical Induction?+
Yes — Class 11 Mathematics Principle of Mathematical Induction has timed online quizzes at Easy, Medium and Hard levels, with instant scoring and a worked explanation on every question. The first quiz on the chapter is free.