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TMUA 2017 · Paper 1 · Question 13 of 20

TMUA 2017 Paper 1 Question 13

Sequences and series — Binomial expansion · comparing two coefficients. Try it first; the answer and a full worked solution are below.

TMUA 2017 · Paper 1Sequences and seriesBinomial expansion · comparing two coefficients6 options
In the expansion of (a + bx)5 the coefficient of x4 is 8 times the coefficient of x2.

Given that a and b are non-zero positive integers, what is the smallest possible value of a + b?

  1. A3
  2. B4
  3. C5
  4. D9
  5. E13
  6. F17
Show the answer and worked solution
answer · C
  1. A3
  2. B4
  3. C5
  4. D9
  5. E13
  6. F17
The term in xk is (5k)a5−kbk xk, so the coefficients of x4 and x2 are 5ab4 and 10a3b2. The condition gives 5ab4 = 80a3b2, and dividing by 5ab2 (both non-zero) leaves b2 = 16a2, so b = 4a for positive integers. The smallest case is a = 1, b = 4, giving a + b = 5.