Binomial Expansion

In the binomial expansion of (1 + x)⁵, find the coefficient of x².

• A5
• B10✓
• C25
• D20

Answer: The coefficient of x² is the binomial coefficient C(5,2) = 10.

Expand (2 + 3x)⁴ and state the coefficient of x².

• A108
• B216✓
• C54
• D144

Answer: The x² term is C(4,2)·2²·(3x)² = 6 × 4 × 9x² = 216x², so the coefficient is 216.

In the expansion of (1 − 2x)⁶, find the coefficient of x³.

• A160
• B−60
• C−160✓
• D−320

Answer: The x³ term is C(6,3)·(−2x)³ = 20 × (−8)x³ = −160x³, so the coefficient is −160. (Forgetting to cube the −2 gives wrong values such as −60.)

Find the term in x² in the expansion of (2x − 1)⁵.

• A40x²
• B−80x²
• C10x²
• D−40x²✓

Answer: The general term is C(5,r)(2x)^(5−r)(−1)^r. For x² we need 5 − r = 2, so r = 3: C(5,3)(2x)²(−1)³ = 10 × 4x² × (−1) = −40x².

Find the first three terms, in ascending powers of x, of the expansion of (1 + x)^(1/2).

• A1 + (1/2)x − (1/8)x²✓
• B1 + (1/2)x + (1/8)x²
• C1 − (1/2)x + (1/8)x²
• D1 + (1/2)x − (1/4)x²

Answer: Using (1 + x)ⁿ with n = 1/2: term 1 = 1; term 2 = (1/2)x; term 3 = [(1/2)(1/2 − 1)/2!]x² = [(1/2)(−1/2)/2]x² = −(1/8)x².

Find the first three terms, in ascending powers of x, of the expansion of (1 − 3x)⁻².

• A1 − 6x + 27x²
• B1 + 6x + 9x²
• C1 + 6x + 27x²✓
• D1 + 6x − 27x²

Answer: Using (1 + y)ⁿ with n = −2 and y = −3x: 1 + (−2)(−3x) + [(−2)(−3)/2!](−3x)² = 1 + 6x + 3 × 9x² = 1 + 6x + 27x².

Use the first three terms of the expansion of (1 + x)⁸ to estimate (1.02)⁸, by taking x = 0.02. Which value is correct to 4 decimal places?

• A1.1600
• B1.1712✓
• C1.0160
• D1.2000

Answer: (1 + x)⁸ ≈ 1 + 8x + C(8,2)x² = 1 + 8x + 28x². With x = 0.02: 1 + 8(0.02) + 28(0.0004) = 1 + 0.16 + 0.0112 = 1.1712.

The expansion of (1 + 4x)^(1/2) as a series in ascending powers of x is valid only for a certain range of x. State this range.

• A|x| < 4
• B|x| < 1
• C|x| < 1/2
• D|x| < 1/4✓

Answer: The expansion of (1 + bx)ⁿ for non-integer n is valid when |bx| < 1. Here b = 4, so |4x| < 1, giving |x| < 1/4.

Find the coefficient of x² in the expansion of (1 + x)⁶.

• A360
• B6
• C12
• D15✓

Answer: The coefficient of x² in (1 + x)^n is nCk = 6C2 = 15. 6C1, the neighbouring term, is offered too — an off-by-one in the counting is the usual cause.

Find the coefficient of x³ in the expansion of (1 + x)⁸.

• A6720
• B28
• C24
• D56✓

Answer: The coefficient of x³ in (1 + x)^n is nCk = 8C3 = 56. 8C2, the neighbouring term, is offered too — an off-by-one in the counting is the usual cause.

Find the coefficient of x² in the expansion of (1 + x)⁵.

• A7
• B5
• C60
• D10✓

Answer: The coefficient of x² in (1 + x)^n is nCk = 5C2 = 10. 5C1, the neighbouring term, is offered too — an off-by-one in the counting is the usual cause.

How many terms are in the expansion of (1 + x)⁷?

• A7, matching the power that the bracket has been raised to
• B8✓
• C6, which is one fewer than the index of the expansion
• D14, since each of the seven terms produces a pair

Answer: The powers of x run from 0 up to 7 inclusive, giving n + 1 = 8 terms. Counting from x⁰ rather than x¹ is the step people miss.

For what values of x is the expansion of (1 + x)⁻² valid?

• AOnly for values of x that are greater than or equal to zero
• BFor every real value of x without any restriction at all
• C|x| < 1✓
• DOnly for whole-number values of x, since the index is negative

Answer: With a negative or fractional index the binomial expansion is an infinite series, and it only converges for |x| < 1. A positive whole-number index gives a finite expansion valid everywhere — the restriction appears precisely because the series never terminates.