TTMUA Lab
TMUA 2023 · Paper 1 · Question 6 of 20

TMUA 2023 Paper 1 Question 6

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

TMUA 2023 · Paper 1Sequences and seriesBinomial expansion · divisibility of coefficients7 optionshard
In the simplified expansion of (2 + 3x)12, how many of the terms have a coefficient that is divisible by 12?
  1. A0
  2. B2
  3. C5
  4. D10
  5. E11
  6. F12
  7. G13
Show the answer and worked solution
answer · E
  1. A0
  2. B2
  3. C5
  4. D10
  5. E11
  6. F12
  7. G13
The expansion has 13 terms, the one in xk having coefficient (12k)212−k3k. Divisibility by 12 needs a factor 4 and a factor 3. For 1 ≤ k ≤ 10 the 3k supplies the 3 and 212−k supplies at least 4, so all ten qualify. For k = 11 the coefficient is 12 × 2 × 311 = 24 × 311, which also qualifies. But k = 0 gives 212, with no factor of 3, and k = 12 gives 312, which is odd. So 11 terms qualify.