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TMUA 2018 · Paper 2 · Question 2 of 20

TMUA 2018 Paper 2 Question 2

Sequences and series — Binomial expansion · picking out the constant term. Try it first; the answer and a full worked solution are below.

TMUA 2018 · Paper 2Sequences and seriesBinomial expansion · picking out the constant term6 options
Find the value of the constant term in the expansion of (x6 − 1x2)12
  1. A−495
  2. B−220
  3. C−66
  4. D66
  5. E220
  6. F495
Show the answer and worked solution
answer · B
  1. A−495
  2. B−220
  3. C−66
  4. D66
  5. E220
  6. F495
The general term is (12k)(x6)12−k(−x−2)k = (12k)(−1)k x72 − 6k − 2k. The power of x is 72 − 8k, which is zero when k = 9. That term is (129)(−1)9 = −220. The sign matters: the negative in the bracket is raised to an odd power, so the answer is negative.