ECAT — Engineering College Admission Test (UET)
Take the ECAT Mathematics — Mathematical Induction and Binomial Theorem (Set 7): 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. The coefficient of x² in the expansion of (1 + 3x)^(2/3), |x| < 1/3, is
Answer: A. −1
Explanation: The coefficient is [(2/3)(−1/3)/2!]·3² = (−2/9)/2·9 = −1.
Q2. Using the first two terms of the binomial series for (1 + x)^(1/3), the value of ∛8.24 is approximately
Answer: D. 2.02
Explanation: ∛8.24 = 2(1 + 0.03)^(1/3) ≈ 2(1 + 0.03/3) = 2(1.01) = 2.02.
Q3. In proving 1/(1·2) + 1/(2·3) + … + 1/(n(n + 1)) = n/(n + 1) by induction, the inductive step adds 1/((k + 1)(k + 2)) to k/(k + 1). Written over the common denominator (k + 1)(k + 2), the numerator is
Answer: C. k² + 2k + 1
Explanation: k/(k + 1) = k(k + 2)/((k + 1)(k + 2)), so the numerator is k(k + 2) + 1 = k² + 2k + 1 = (k + 1)². Dividing by (k + 1)(k + 2) gives (k + 1)/(k + 2).
Q4. The inequality n! > 3ⁿ is false for small n. The smallest positive integer n₀ from which it holds for every n ≥ n₀ is
Answer: A. 7
Explanation: 6! = 720 < 3⁶ = 729 (false) but 7! = 5040 > 3⁷ = 2187. For n ≥ 7, going from n to n + 1 multiplies the left side by n + 1 ≥ 8 > 3, so the inequality persists. n₀ = 7.
Q5. The largest integer that divides 4ⁿ + 15n − 1 for every positive integer n is
Answer: C. 9
Explanation: n = 1 gives 18 and n = 2 gives 45, whose gcd is 9. In general 4ⁿ = (1 + 3)ⁿ = 1 + 3n + 9(…), so 4ⁿ + 15n − 1 = 18n + 9(…), a multiple of 9. The largest guaranteed divisor is 9.
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Q6. The largest integer that divides 2ⁿ⁺²·3ⁿ + 5n − 4 for every positive integer n is
Q7. The remainder when 2¹⁰⁰ is divided by 9 is
Q8. The coefficient of x⁶ in the expansion of (x² + 3/x)⁶ is
Q9. In the expansion of (2 + 3x)⁹ the coefficients of xʳ and xʳ⁺¹ are equal. The value of r is
Q10. The 4th term from the end in the expansion of (a + b)¹⁰ is
Q11. In the expansion of 1/((1 − x)(1 − x²)) in ascending powers of x, the coefficient of x⁶ is
Q12. The binomial series of √(1 + 4x) in ascending powers of x is valid for
Q13. The value of (√5 + 2)⁵ − (√5 − 2)⁵ is
Q14. Using the identity 1³ + 3³ + … + (2n − 1)³ = n²(2n² − 1), the value of 1³ + 3³ + 5³ + 7³ + 9³ is
Q15. Pascal's triangle starts with row 0 = (1). The total of all the entries in rows 0 to 7 is
Q16. The value of C(2, 2) + C(3, 2) + C(4, 2) + … + C(10, 2) is
Q17. The value of Σ r(r − 1)·C(6, r), summed over r = 0 to 6, is
Q18. Using the first three terms of the binomial series, (1.03)⁻² is approximately
Q19. In the expansion of (1 + 2x)⁸, the ratio of the coefficient of x³ to the coefficient of x² is
Q20. In a standard 'proof' that all horses in any group of n horses have the same colour, the inductive step from P(k) to P(k + 1) breaks down for one particular value of k. That value is
Q21. In the expansion of (x + a)ⁿ the 2nd, 3rd and 4th terms are 240, 720 and 1080 respectively. Then n equals
Q22. The last two digits of 11¹² are
Q23. In the expansion of (x + 1/x)⁶, the sum of the coefficients of x² and x⁻² is
Q24. The sum of the coefficients of x² and x³ in the expansion of (1 + 2x)⁵ is
Q25. The term independent of x in the expansion of (x⁴ + 1/x)¹⁰ is
Q26. The coefficients of x⁰, x¹, …, x¹⁰ in the expansion of (1 − x)⁻² are added. The total is
Q27. The number of terms in the expansion of (1 + 2x)⁶(1 − 2x)⁶ after simplification is
Q28. The value of C(4,0)C(6,3) + C(4,1)C(6,2) + C(4,2)C(6,1) + C(4,3)C(6,0) is
Q29. In the expansion of (1 + x)¹⁰, the ratio T₄/T₃ of the 4th term to the 3rd term, at x = 1/2, is
Q30. The sum of the binomial coefficients in the expansion of (1 + x)ⁿ is 4096. The coefficient of x³ in that expansion is