O-Level Binomial expansions: worked solution

Question

In the expansion of (3 + x/2)ⁿ the x² coefficient is triple the x³ coefficient. Find n.

Step 1: write the general term of (3 + x/2)ⁿ. By the binomial theorem, the term containing x^r is C(n, r) × 3^(n−r) × (x/2)^r = C(n, r) × 3^(n−r) × x^r / 2^r. So the coefficient of x^r is C(n, r) × 3^(n−r) / 2^r. The 1/2^r must not be forgotten. Step 2: write down the two coefficients the question compares. coefficient of x² = C(n, 2) × 3^(n−2) / 2² = C(n, 2) × 3^(n−2) / 4 coefficient of x³ = C(n, 3) × 3^(n−3) / 2³ = C(n, 3) × 3^(n−3) / 8 Step 3: turn the given condition into an equation. "The x² coefficient is triple the x³ coefficient" means C(n, 2) × 3^(n−2) / 4 = 3 × C(n, 3) × 3^(n−3) / 8. On the right-hand side, 3 × 3^(n−3) = 3^(n−2), so C(n, 2) × 3^(n−2) / 4 = C(n, 3) × 3^(n−2) / 8. Step 4: cancel and tidy. The factor 3^(n−2) appears on both sides and is never zero, so divide both sides by it: C(n, 2) / 4 = C(n, 3) / 8. Multiply both sides by 8: 2 C(n, 2) = C(n, 3). Step 5: write the binomial coefficients in terms of n and solve. C(n, 2) = n(n − 1)/2 and C(n, 3) = n(n − 1)(n − 2)/6, so 2 × n(n − 1)/2 = n(n − 1)(n − 2)/6 n(n − 1) = n(n − 1)(n − 2)/6. For there to be a term in x³ at all, n must be at least 3, so n(n − 1) is not zero and both sides can be divided by it: 1 = (n − 2)/6 6 = n − 2 n = 8. Check with n = 8: coefficient of x² = C(8, 2) × 3⁶/4 = 28 × 729/4 = 5103, coefficient of x³ = C(8, 3) × 3⁵/8 = 56 × 243/8 = 1701, and 3 × 1701 = 5103, as required.