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TMUA 2021 · Paper 1 · Question 16 of 20

TMUA 2021 Paper 1 Question 16

Sequences and series — Binomial expansion · ascending and descending terms. Try it first; the answer and a full worked solution are below.

TMUA 2021 · Paper 1Sequences and seriesBinomial expansion · ascending and descending terms5 optionshard
Consider the expansion of (a + bx)n The third term, in ascending powers of x, is 105x2

The fourth term, in ascending powers of x, is 210x3

The fourth term, in descending powers of x, is 210x3

Find the value of (ab)2

  1. A14
  2. B49
  3. C2536
  4. D56
  5. E1
Show the answer and worked solution
answer · B
  1. A14
  2. B49
  3. C2536
  4. D56
  5. E1
The fourth term in descending powers carries xn−3, and it is given as a term in x3, so n − 3 = 3 and n = 6. Now the ascending conditions read (62)a4b2 = 15a4b2 = 105 and (63)a3b3 = 20a3b3 = 210, so a4b2 = 7 and a3b3 = 212. Dividing one by the other gives ab = 721/2 = 23, so (ab)2 = 49. Note the third condition is what pins down n; without it n is unknown.