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TMUA 2016 · Paper 1 · Question 19 of 20

TMUA 2016 Paper 1 Question 19

Sequences and series — Coefficients from a trinomial power and a binomial expansion. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 1Sequences and seriesCoefficients from a trinomial power and a binomial expansion5 optionshard
The coefficient of x3 in the expansion of (1 + 2x + 3x2)6 is equal to twice the coefficient of x4 in the expansion of (1 − ax2)5.

Find all possible values of the constant a.

  1. A± 2√2
  2. B±√17
  3. C±√34
  4. D± 2√17
  5. EThere are no possible values of a.
Show the answer and worked solution
answer · B
  1. A± 2√2
  2. B±√17
  3. C±√34
  4. D± 2√17
  5. EThere are no possible values of a.
For the trinomial, ask which choices of 2x and 3x2 from the six brackets give x3. Taking three 2x terms: (63)(2x)3 = 20 ⋅ 8x3 = 160x3. Taking one 2x and one 3x2: there are 6 × 5 = 30 ordered ways to pick the two distinct brackets, contributing 30 ⋅ 2 ⋅ 3 = 180. So the coefficient of x3 is 340. For (1 − ax2)5, an x4 term needs (−ax2)2, giving (52)a2 = 10a2. Then 340 = 2 × 10a2, so a2 = 17 and a = ±√17.