IMO Practice Test: Principle of Mathematical Induction | Class 11 Mathematics

Question 1 of 6 hard

A student "proves" that every positive integer equals its successor by assuming n = n+1 and deriving (n+1) = (n+2). The fatal flaw in this argument is that:

the inductive step is wrong

there is no true base case, so the chain never starts

induction does not apply to integers

the hypothesis should be n = n−1

Explanation: The statement n = n+1 is false for every n, so no base case can hold. Without a valid base case the induction proves nothing, however the step is manipulated.

Question 2 of 6 hard

Using 1³ + 2³ + ... + n³ = [n(n+1)/2]², the value of 1³ + 2³ + ... + 10³ is:

2025

3025

1225

55

Explanation: The sum equals [10·11/2]² = 55² = 3025.

Question 3 of 6 hard

For all n ∈ ℕ, the expression 10ⁿ + 3·4^(n+2) + 5 is divisible by:

7

9

11

13

Explanation: This standard NCERT result is divisible by 9; an induction (or modular) check confirms 9 | (10ⁿ + 3·4^(n+2) + 5) for every n.

Question 4 of 6 hard

In proving Bernoulli’s inequality (1+x)ⁿ ≥ 1 + nx for x > −1, the inductive step relies crucially on the fact that:

1 + x > 0, so multiplying the hypothesis by it preserves the inequality

x is an integer

n is even

x² = 1

Explanation: Because 1 + x > 0, multiplying (1+x)^k ≥ 1 + kx by (1+x) keeps the inequality direction; the dropped term kx² ≥ 0 then gives the result.

Question 5 of 6 hard

The smallest natural number n for which the statement "n! > 2ⁿ" is true is:

2

3

4

5

Explanation: Compare: 3! = 6 < 8 = 2³ (false), but 4! = 24 > 16 = 2⁴ (true). So n! > 2ⁿ first holds at n = 4.

Question 6 of 6 hard

For all n ∈ ℕ, 7^(2n) + 2^(3n−3)·3^(n−1) is divisible by:

25

36

125

11

Explanation: This classic HOTS result is divisible by 25: with the rewrite 8^(n−1)·3^(n−1) = 24^(n−1), induction shows 25 | (49ⁿ + ⋯), so the divisor is 25.

IMO Practice Test — Principle of Mathematical Induction