Pure Mathematics — First: Algebra · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 37 questions on this lesson, each with an explanation.
How many terms are there in the expansion of (2x - 5)²⁰?
Correct answer: (D) 21
Explanation: Step 1: The expansion of (a + b)ⁿ has n + 1 terms. Step 2: Here n = 20, so number of terms = 20 + 1 = 21. ∴ The expansion has 21 terms.
In the expansion of (a + b)ⁿ, the general term Tᵣ₊₁ is equal to which of the following?
Correct answer: (D) ⁿCᵣ aⁿ⁻ʳ bʳ
Explanation: Step 1: The general term of (a + b)ⁿ is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ, for r = 0, 1, 2, ..., n. Step 2: The power of a decreases from n while the power of b increases from 0, and the two powers add to n. ∴ Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ.
What is the last term in the expansion of (x + 2)⁶?
Correct answer: (C) 64
Explanation: Step 1: The last term is Tₙ₊₁ = ⁿCₙ x⁰ aⁿ. Step 2: For (x + 2)⁶, last term = ⁶C₆ · 2⁶ = 1 · 64. ∴ The last term is 64.
In the expansion of (x/2 + 3/x)¹¹, the two middle terms are equal. Find x² (x ≠ 0).
Correct answer: (D) 6
Explanation: Step 1: n = 11 is odd, so the two middle terms are T₆ and T₇. Step 2: T₆ = ¹¹C₅ (x/2)⁶ (3/x)⁵ and T₇ = ¹¹C₆ (x/2)⁵ (3/x)⁶, and ¹¹C₅ = ¹¹C₆. Step 3: T₆ = T₇ gives x/2 = 3/x. Step 4: x² = 6.
In the expansion of (2x − 1/x)⁷ according to the descending powers of x, find the fourth term T₄.
Correct answer: (D) −560x
Explanation: Step 1: T₄ = T₃₊₁, so r = 3: T₄ = ⁷C₃ (2x)⁴ (−1/x)³. Step 2: ⁷C₃ = 35, (2x)⁴ = 16x⁴, (−1/x)³ = −1/x³. Step 3: T₄ = 35 × 16 × (−1) x⁴⁻³ = −560x.
Find in the simplest form: (√3 + 1)⁵ − (√3 − 1)⁵
Correct answer: (D) 152
Explanation: Step 1: (x + a)ⁿ − (x − a)ⁿ = 2(T₂ + T₄ + T₆ + …) with x = √3, a = 1, n = 5. Step 2: T₂ = ⁵C₁ (√3)⁴ = 5 × 9 = 45, T₄ = ⁵C₃ (√3)² = 10 × 3 = 30, T₆ = ⁵C₅ = 1. Step 3: The expression = 2(45 + 30 + 1) = 152.
In the expansion of (1 + x)ⁿ, the coefficient of the third term is 105. Find n.
Correct answer: (B) 15
Explanation: Step 1: The 3rd term of (1 + x)ⁿ has coefficient ⁿC₂. Step 2: ⁿC₂ = 105 gives n(n - 1)/2 = 105, so n(n - 1) = 210. Step 3: 15 · 14 = 210, so n = 15. ∴ n = 15.
If (1 + cx)ⁿ = 1 + 24x + a₁x² + a₂x³ + … + cⁿxⁿ and 3a₂ = 16a₁, find n and c (c ≠ 0).
Correct answer: (B) n = 6, c = 4
Explanation: Step 1: The coefficient of x is ⁿC₁ c = 24, so nc = 24 … (1) Step 2: a₁ = ⁿC₂ c² and a₂ = ⁿC₃ c³, so a₂/a₁ = (ⁿC₃/ⁿC₂) × c = (n − 2)c/3. Step 3: 3a₂ = 16a₁ gives a₂/a₁ = 16/3, so (n − 2)c = 16, i.e. nc − 2c = 16. Step 4: From (1): 24 − 2c = 16, so c = 4 and n = 24 ÷ 4 = 6.
What is the highest coefficient in the expansion of (1 + x)¹⁴?
Correct answer: (C) 3432
Explanation: Step 1: In (1 + x)¹⁴ the binomial coefficients are ¹⁴Cᵣ; the highest is the middle one at r = 7. Step 2: Highest coefficient = ¹⁴C₇. Step 3: ¹⁴C₇ = 3432. ∴ The highest coefficient is 3432.
If ⁿC₂ = ⁿC₆, find the value of n.
Correct answer: (D) 8
Explanation: Step 1: ⁿCᵣ = ⁿCₙ₋ᵣ; we are given ⁿC₂ = ⁿC₆ with 2 not equal to 6. Step 2: So the lower parts must add to n: 2 + 6 = n. Step 3: n = 8. ∴ n = 8.
The first 10 questions in every subject are free, each with an explanation.