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O-Level Binomial expansions
What the O-Level syllabus expects for Binomial expansions, and how to practise it.
What the syllabus expects
- Applying the Binomial Theorem when n is a positive integer
- Using the notation n! together with the binomial coefficient notation
- Applying the general term for 0 <= r <= n
Scope: Candidates need not know the greatest term or the properties of the coefficients
How it's examined
Questions on this topic most often ask you to find, compare, determine, solve. About 4% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (4 marks)
The product (1 − 2x − kx²)(2 − 3/x)⁸ contains no term in x. Use this to find the constant term of the expansion.
Show the worked answer
General term of (2 - 3/x)⁸: C(8,r)·2^(8-r)·(-3)^r·x^(-r), giving powers x⁰ down to x⁻⁸. Multiply by (1 - 2x - kx²). Term in x: from -2x × (x⁰ term r=0) = -2x×256 = -512x; from -kx² × (x⁻¹ term r=1) = -kx² × [C(8,1)2⁷(-3)] = -kx²×(-3072x⁻¹) = 3072k x. No term in x: -512 + 3072k = 0 → k = 1/6. Constant term (x⁰): from 1×(x⁰, r=0)=256; from -2x×(x⁻¹, r=1)=-2×(-3072)=6144; from -kx²×(x⁻², r=2)= -k×[C(8,2)2⁶(-3)²] = -k×16128 = -16128×(1/6) = -2688. Constant = 256 + 6144 - 2688 = 3712.
Example 2 (5 marks)
The product (2 + x²)(1/(2x) - ax)^5 turns out to have no term in x. Assuming a is non-zero, work out the constant a.
Show the worked answer
Expand (1/(2x) - ax)⁵. General term = C(5,k)(1/(2x))^(5-k)(-ax)^k = C(5,k)(1/2)^(5-k)(-a)^k x^(2k-5). So the bracket contains powers x^(2k-5): x⁻⁵, x⁻³, x⁻¹, x¹, x³, x⁵. Multiplying by (2 + x²), a term in x¹ arises two ways: (i) 2 x (x¹ term of bracket): need 2k-5=1 -> k=3: C(5,3)(1/2)²(-a)³ = 10*(1/4)*(-a³) = -(5/2)a³; times 2 = -5a³. (ii) x² x (x⁻¹ term of bracket): need 2k-5=-1 -> k=2: C(5,2)(1/2)³(-a)² = 10*(1/8)*a² = (5/4)a². No term in x => -5a³ + (5/4)a² = 0 => (5/4)a²(1 - 4a) = 0. Since a != 0, a = 1/4.
Example 3 (4 marks)
Expand (2 + x²/2)^5 as far as the first three terms in ascending powers of x and simplify them, then apply the result to approximate (2.005)^5, setting out all your working.
Show the worked answer
(2 + x²/2)⁵, first three terms in ascending powers of x: Term 1: C(5,0)·2⁵ = 32 Term 2: C(5,1)·2⁴·(x²/2) = 5·16·(x²/2) = 40x² Term 3: C(5,2)·2³·(x²/2)² = 10·8·(x⁴/4) = 20x⁴ So (2 + x²/2)⁵ ≈ 32 + 40x² + 20x⁴. For (2.005)⁵, set 2 + x²/2 = 2.005, so x²/2 = 0.005, x² = 0.01, x = 0.1. Approximation = 32 + 40(0.01) + 20(0.0001) = 32 + 0.4 + 0.002 = 32.402.
More worked questions on this topic
- Expand (2 − 3/x)⁸ and write, in simplified form, its first three terms in descending powers of (2 marks)
- Using that k, show there is no constant term in (1+x⁴)(x−k/x³)^8. (4 marks)
- In the expansion of (3 + x/2)ⁿ the x² coefficient is triple the x³ coefficient. Find n. (4 marks)
- Using (1 + 3x)^8 = 1 + 24x + 252x² + ..., deduce the first three terms of (2 + 6x + 3x²)^8. (3 marks)
- Taking n = 7, demonstrate that the product (48/x² − x⁴)(√x − 2/x)⁷ contains no constant term. (5 marks)
- Find, in ascending powers of x, the first four terms of the expansion of (a/x + 3x²)⁶, simplify (3 marks)
- Taking n = 8, and given that the fifth term of (px − q/x)⁸ equals 1120 with p − q = 1, determin (5 marks)
- In the expansion of (2+x)^n (n a positive integer) the x² coefficient is double the x coefficie (3 marks)
- The expansion of (1 + ax)^n starts 1 - 3x + (33/8)x², where a and n are constants. Determine th (5 marks)
More O-Level Additional Maths (A-Maths) topics
Quadratic functions · Equations and inequalities · Surds · Polynomials and partial fractions · Exponential and logarithmic functions · Trigonometric functions, identities and equations · all of O-Level Additional Maths (A-Maths)