ECAT — Engineering College Admission Test (UET)
Take the ECAT Mathematics — Mathematical Induction and Binomial Theorem (Set 1): 30 multiple-choice questions (40 minutes) with instant scoring and a full…
What this test covers
The first 5 questions of this test, with the correct answer and explanation.
Q1. In a proof by mathematical induction for statements P(n), n ∈ ℕ, the first step is to:
Answer: A. verify that P(1) is true
Explanation: The base case P(1) is checked first; then P(k) ⇒ P(k + 1) is proved (the inductive step).
Q2. A statement P(n) is true for all n ∈ ℕ by mathematical induction if:
Answer: B. P(1) is true and P(k) true implies P(k + 1) true
Explanation: Both conditions are needed: the base case and the inductive step P(k) ⇒ P(k + 1). Checking finitely many cases is not a proof.
Q3. The smallest positive integer n₀ such that 2ⁿ > n² holds for every integer n ≥ n₀ is:
Answer: B. 5
Explanation: 2ⁿ > n² holds at n = 1 but fails at n = 2, 3 (4 = 4, 8 < 9) and n = 4 (16 = 16). From n = 5 (32 > 25) it holds for all larger n, which can be proved by induction from base case 5.
Q4. For every positive integer n, n³ − n is divisible by:
Answer: D. 6
Explanation: n³ − n = (n − 1)n(n + 1) is a product of three consecutive integers, so it is divisible by 2 and by 3, hence by 6. (For n = 2 it equals 6, which is not divisible by 4, 5 or 9.)
Q5. For every positive integer n, 5ⁿ − 1 is divisible by:
Answer: C. 4
Explanation: 5ⁿ − 1 = (1 + 4)ⁿ − 1 is a sum of multiples of 4; e.g. 5 − 1 = 4, 25 − 1 = 24. Induction: 5ᵏ⁺¹ − 1 = 5(5ᵏ − 1) + 4.
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Q6. To prove 1 + 3 + 5 + … + (2n − 1) = n² by induction, assume it holds for n = k. The term added to both sides to obtain the case n = k + 1 is:
Q7. The number of terms in the expansion of (a + b)¹⁰ is:
Q8. The coefficient of x³ in the expansion of (1 + x)⁷ is:
Q9. The middle term in the expansion of (x + 1/x)⁸ is:
Q10. The term independent of x in the expansion of (x² + 1/x)⁶ is:
Q11. The sum of all the binomial coefficients in the expansion of (1 + x)⁸ is:
Q12. In the expansion of (1 + x)ⁿ, the sum of the coefficients of the odd-numbered terms (1st, 3rd, 5th, …) is:
Q13. The general term Tᵣ₊₁ in the expansion of (a + b)ⁿ is:
Q14. The 4th term in the expansion of (2x − 3)⁵ is:
Q15. The coefficient of x⁴ in the expansion of (1 − 2x)⁶ is:
Q16. The binomial expansion (1 + x)⁻¹ = 1 − x + x² − x³ + … is valid only when:
Q17. The coefficient of x² in the expansion of (1 − x)⁻² is:
Q18. Using the first two terms of the binomial expansion, (1.002)¹⁰ is approximately:
Q19. In the expansion of (1 + x)^(1/2), the coefficient of x² is:
Q20. The value of ¹⁰C₀ − ¹⁰C₁ + ¹⁰C₂ − … + ¹⁰C₁₀ is:
Q21. The value of (√2 + 1)⁴ + (√2 − 1)⁴ is:
Q22. The coefficient of x⁵ in the expansion of (x + 2)⁷ is:
Q23. The expansion of (a + b)⁹ has:
Q24. The binomial coefficients in the expansion of (a + b)⁵, as shown in Pascal's triangle, are:
Q25. If the coefficients of x² and x³ in the expansion of (1 + x)ⁿ are equal, then n equals:
Q26. The expansion of (1 − 2x)^(−1/2) in ascending powers of x is valid for:
Q27. The coefficient of x³ in the expansion of (1 + x)⁻³ is:
Q28. Using the binomial theorem, 11¹⁰ − 1 is divisible by:
Q29. To prove 1 + 2 + 4 + … + 2ⁿ⁻¹ = 2ⁿ − 1 by induction, after assuming the result for n = k one must show that:
Q30. The expression n² − n + 41 is prime for n = 1, 2, …, 40. For n = 41 its value is: