Online Test: Principle of Mathematical Induction | Class 11 Mathematics

Question 1 of 10 easy

A proof by the principle of mathematical induction requires which two parts?

Two base cases

A base case and an inductive step

Only an inductive step

Checking P(1), P(2) and P(3)

Explanation: Induction needs the base case P(1) and the inductive step P(k) ⇒ P(k+1).

Question 2 of 10 easy

For a statement claimed to be true for all n ≥ 4, the base case to be verified is:

P(1)

P(0)

P(4)

P(5)

Explanation: The base case is taken at the smallest asserted value, n₀ = 4, i.e. P(4).

Question 3 of 10 easy

The sum 1 + 2 + 3 + ... + n equals:

n(n+1)

n(n+1)/2

n(n+1)(2n+1)/6

2ⁿ − 1

Explanation: The standard formula, provable by induction, is n(n+1)/2.

Question 4 of 10 easy

In an inductive proof, the assumption that the statement holds for n = k is called the:

Conclusion

Base case

Inductive hypothesis

Converse

Explanation: Assuming P(k) is the inductive hypothesis, used to derive P(k+1).

Question 5 of 10 medium

For all natural numbers n, the expression 3²ⁿ − 1 is always divisible by:

3

6

8

9

Explanation: 3^(2(k+1)) − 1 = 9(8m + 1) − 1 = 72m + 8 = 8(9m + 1), so it is divisible by 8.

Question 6 of 10 medium

The sum of the first n squares, 1² + 2² + ... + n², equals:

n(n+1)/2

[n(n+1)/2]²

n(n+1)(2n+1)/6

n³

Explanation: The closed form is n(n+1)(2n+1)/6.

Question 7 of 10 medium

Why is verifying P(1), P(2), ..., P(50) not a proof that P(n) holds for every n?

Because P(1) might be false

Because finitely many checks cannot cover all natural numbers; the inductive step is required

Because you must also check P(0)

Because the cases are not independent

Explanation: A property can hold for the first several values and then fail (e.g. n² − n + 41 is prime for n ≤ 40 but not at n = 41). Only the inductive step makes the conclusion universal.

Question 8 of 10 medium

For all n ∈ ℕ, the number n³ − n is divisible by:

4

5

6

7

Explanation: (k+1)³ − (k+1) = (k³ − k) + 3k(k+1); since k(k+1) is even, 3k(k+1) is a multiple of 6, so the whole expression is divisible by 6.

Question 9 of 10 hard

The inequality 2ⁿ > n² is true for:

all n ≥ 1

all n ≥ 2

all n ≥ 5

no value of n

Explanation: 2ⁿ > n² fails at n = 2, 3, 4 but holds for all n ≥ 5 (32 > 25), which is the base case.

Question 10 of 10 medium

In the inductive step of a divisibility proof, f(k+1) is typically written as:

f(k) × d

f(k) + (a multiple of d)

d − f(k)

f(k) − 1

Explanation: Writing f(k+1) = f(k) + (multiple of d) lets you conclude d | f(k+1), since both summands are divisible by d.

Online Test — Principle of Mathematical Induction