O-Level Binomial expansions: worked solution

Question

The expansion of (1 + ax)^n starts 1 - 3x + (33/8)x², where a and n are constants. Determine the values of a and n. [5]

Write out the first three terms of the binomial expansion, then compare coefficients with the given expansion. For (1 + y)^n the expansion begins 1 + ny + [n(n − 1)/2]y² + ... Here y = ax, so (1 + ax)^n = 1 + n(ax) + [n(n − 1)/2](ax)² + ... = 1 + nax + [n(n − 1)/2]a²x² + ... Compare this with the given start 1 − 3x + (33/8)x². Comparing the coefficients of x: na = −3 ...(1) Comparing the coefficients of x²: n(n − 1)a²/2 = 33/8 ...(2) There are two unknowns and two equations. Eliminate a: from (1), a = −3/n. Substitute a = −3/n into (2): n(n − 1)(−3/n)²/2 = 33/8 (−3/n)² = 9/n², so n(n − 1)(9/n²)/2 = 33/8 One factor of n cancels: 9(n − 1)/(2n) = 33/8. Cross-multiply: 8 × 9(n − 1) = 33 × 2n 72(n − 1) = 66n 72n − 72 = 66n 72n − 66n = 72 6n = 72 n = 12. Substitute n = 12 back into (1): 12a = −3 a = −3/12 = −1/4. Check the x² coefficient: n(n − 1)a²/2 = 12 × 11 × (1/16) ÷ 2 = 132/32 = 33/8, as required. So a = −1/4 and n = 12.