O-Level Binomial expansions: worked solution

Question

Taking n = 8, and given that the fifth term of (px − q/x)⁸ equals 1120 with p − q = 1, determine p and q. [5]

Step 1: write the general term of the binomial expansion. For (a + b)ⁿ the general term is C(n, r)a^(n−r)b^r, so here with a = px, b = −q/x and n = 8: T = C(8, r) (px)^(8−r) (−q/x)^r Step 2: work out which value of r gives the fifth term. The terms run r = 0 for the first term, r = 1 for the second, and so on, so r is always one less than the term number. Fifth term means r = 4. Step 3: substitute r = 4 and simplify. T₅ = C(8, 4) (px)⁴ (−q/x)⁴ C(8, 4) = 8!/(4!4!) = (8×7×6×5)/(4×3×2×1) = 70 (px)⁴ = p⁴x⁴ (−q/x)⁴ = q⁴/x⁴ - the fourth power of a negative quantity is positive, since the minus sign is used an even number of times. T₅ = 70 × p⁴x⁴ × q⁴/x⁴ The x⁴ cancels with the x⁴ in the denominator, which is why the fifth term is the constant one: T₅ = 70p⁴q⁴ = 70(pq)⁴ Step 4: use the given value of the fifth term. 70(pq)⁴ = 1120 Divide both sides by 70: (pq)⁴ = 16 Take the fourth root of both sides. Since 2⁴ = 16: pq = 2 Step 5: solve pq = 2 together with the given p − q = 1. From p − q = 1, rearrange to p = q + 1. Substitute into pq = 2: (q + 1)q = 2 q² + q − 2 = 0 Factorise: two numbers multiplying to −2 and adding to 1 are +2 and −1. (q + 2)(q − 1) = 0 q = −2 or q = 1 Taking the positive constants, q = 1, and then p = q + 1 = 2. Check: pq = 2 × 1 = 2 and p − q = 2 − 1 = 1, and the fifth term is 70 × 2⁴ = 70 × 16 = 1120. So p = 2 and q = 1.