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TMUA 2016 · Paper 2 · Question 6 of 20

TMUA 2016 Paper 2 Question 6

Sequences and series — Recursively defined functions · summing a geometric series. Try it first; the answer and a full worked solution are below.

TMUA 2016 · Paper 2Sequences and seriesRecursively defined functions · summing a geometric series8 options
The sequence of functions f1(x), f2(x), f3(x), … is defined as follows:

f1(x) = x10 fn+1(x) = x fn'(x)   for  n ≥ 1

where fn'(x) = dfn(x)dx.

Find the value of ∑n=120 fn(x)

  1. Ax10(x20−1)x−1
  2. Bx10(x21−1)x−1
  3. C(1020−19)x10
  4. D(1021−19)x10
  5. E((10x)20−110x−1)x10
  6. F((10x)21−110x−1)x10
  7. Gx10 + x9 + x8 + ⋯ + x + 1
  8. Hx10 + 10x9 + (10× 9)x8 + ⋯ + (10× 9× ⋯ × 2)x + (10× 9×⋯× 2× 1)
Show the answer and worked solution
answer · C
  1. Ax10(x20−1)x−1
  2. Bx10(x21−1)x−1
  3. C(1020−19)x10
  4. D(1021−19)x10
  5. E((10x)20−110x−1)x10
  6. F((10x)21−110x−1)x10
  7. Gx10 + x9 + x8 + ⋯ + x + 1
  8. Hx10 + 10x9 + (10× 9)x8 + ⋯ + (10× 9× ⋯ × 2)x + (10× 9×⋯× 2× 1)
Work out the first couple of terms and the pattern is immediate. f2(x) = x⋅ 10x9 = 10x10, and f3(x) = x⋅ 100x9 = 100x10. In general the power of x never changes: if fn(x) = c x10 then fn+1(x) = x⋅ 10c x9 = 10c x10, so fn(x) = 10n−1x10. Summing the geometric series of coefficients, ∑n=120 10n−1 = 1020−19, and the answer is (1020−19)x10.