Principle of Mathematical Induction — Class 11 MCQs with Answers

Principle of Mathematical Induction — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   First step in PMI is:

   AConcludeBAssume P(k)CProve P(k+1)DVerify base case (n=1)
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   Answer: 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.

2. Q2 of 6Easy

   Sum of first n natural numbers:

   An(n−1)/2Bn²Cn(n+1)/2Dn+1
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   Answer: 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.

3. Q3 of 6Medium

   Sum 1 + 2 + ... + 100 equals:

   A10000B100C5050D505
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   Answer: 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.

4. Q4 of 6Medium

   By induction, 3 divides (5ⁿ − 2ⁿ) for all n ≥ 1. Base case n = 1:

   ACannot verifyB3 = 0C5 − 2 = 3, divisibleDAlways false
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   Answer: 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.

5. Q5 of 6Hard

   Prove 2ⁿ ≥ n + 1 for n ≥ 1. Base n = 1: 2 ≥ 2 ✓. Inductive: assume 2ᵏ ≥ k + 1. Show 2ᵏ⁺¹ ≥ k + 2.

   AInductive step failsBCannot proveC2ᵏ⁺¹ = 2ᵏ + 1D2ᵏ⁺¹ = 2·2ᵏ ≥ 2(k+1) ≥ k+2 for k ≥ 0
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   Answer: 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.

6. Q6 of 6Hard

   n(n²+5) is divisible by:

   A6B5C10D7
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   Answer: 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.

Principle of Mathematical Induction notes: key concepts (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.

Why induction works

ℕ has a least element and no gaps, so starting at 1 and always stepping to the next integer eventually reaches every natural number n.

Sum-of-series proofs

Formulae like 1+2+…+n = n(n+1)/2 are classic targets: assume it for k, add the (k+1)-th term, then simplify to the k+1 form.

Divisibility proofs

To show an expression is divisible by d, rewrite the P(k+1) expression so both a factor of d and a multiple coming from P(k) appear.

Inequality proofs

For results like 2ⁿ > n, the inductive step adds or multiplies carefully, combining P(k) with a true bound to reach the P(k+1) inequality.

Choosing the base value

The base need not be 1; some statements start at n=0 or a larger n₀, and induction then proves P(n) for all n ≥ n₀.

Common pitfall

Skipping the base case, or assuming what you must prove, invalidates the argument. Checking a few cases is evidence, never a proof.

Structure of the proof

State P(n), verify the base, assume P(k), prove P(k+1), then conclude by the principle. Each stage should be written out explicitly.