O-Level Binomial expansions: worked solution

Question

In the expansion of (2+x)^n (n a positive integer) the x² coefficient is double the x coefficient. Find n. [3]

Write down the general term of the binomial expansion of (2 + x)^n. Taking a = 2 and b = x, the term containing x^r is C(n, r) × 2^(n−r) × x^r. Coefficient of x (take r = 1): C(n, 1) × 2^(n−1) = n × 2^(n−1). Coefficient of x² (take r = 2): C(n, 2) × 2^(n−2) = [n(n − 1)/2] × 2^(n−2). The question says the x² coefficient is double the x coefficient, so [n(n − 1)/2] × 2^(n−2) = 2 × n × 2^(n−1). Divide both sides by n × 2^(n−2). This is allowed because n is a positive integer, so n ≠ 0 and 2^(n−2) ≠ 0. Left side becomes (n − 1)/2. Right side becomes 2 × 2^(n−1) ÷ 2^(n−2) = 2 × 2^(n−1−(n−2)) = 2 × 2¹ = 4. So (n − 1)/2 = 4. Multiply both sides by 2: n − 1 = 8 n = 9. Check: coefficient of x = 9 × 2⁸ = 2304, coefficient of x² = 36 × 2⁷ = 4608 = 2 × 2304. ✓