Multiple choice

The sum to $2n$ terms of the series $\displaystyle \log _{3}1-\log _{3}3+\log _{3}9-\log _{3}27....$ is

  1. $\displaystyle \frac{\left ( 2n-1 \right )}{\left ( n-1 \right )}$
  2. $\displaystyle \frac{\left ( n-1 \right )}{\left ( 2n-1 \right )}$
  3. $1 - n$
  4. $-n$
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D Correct answer
Explanation

The series is log3(1) - log3(3) + log3(9) - log3(27) + ... which simplifies to 0 - 1 + 2 - 3 + ... . This is an alternating sum of integers. The sum of the first 2n terms is (0-1) + (2-3) + ... + ((2n-2) - (2n-1)), which is -1 + -1 + ... + -1 (n times). Thus, the sum is -n.

AI explanation

Apply the quotient property of logarithms, log(a) - log(b) = log(a/b), to group the terms into consecutive pairs. The first pair is log_3(1) - log_3(3) = log_3(1/3) = -1, and the second pair is log_3(9) - log_3(27) = log_3(9/27) = log_3(1/3) = -1. Since there are 2n terms in total, this forms exactly n pairs, each evaluating to -1. Multiplying n pairs by -1 gives the total sum as -n.

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