TTMUA Lab
TMUA 2022 · Paper 1 · Question 11 of 20

TMUA 2022 Paper 1 Question 11

Sequences and series — Series of logarithms. Try it first; the answer and a full worked solution are below.

TMUA 2022 · Paper 1Sequences and seriesSeries of logarithms8 options
Evaluate ∑n=1100log10(3 1−n)
  1. A−4950log103
  2. B4950log103
  3. C−5050log103
  4. D5050log103
  5. E1 − 4950log103
  6. F1 + 4950log103
  7. G1 − 5050log103
  8. H1 + 5050log103
Show the answer and worked solution
answer · A
  1. A−4950log103
  2. B4950log103
  3. C−5050log103
  4. D5050log103
  5. E1 − 4950log103
  6. F1 + 4950log103
  7. G1 − 5050log103
  8. H1 + 5050log103
Bring the exponent down: log10(31−n) = (1−n)log103. Summing, ∑n=1100(1−n) = 100 − 100 × 1012 = 100 − 5050 = −4950. So the total is −4950log103.