Which one of the following is a geometric series?

Question

BlackTom AI · Accepted Answer

A geometric series has terms a_n that form a constant multiplicative ratio r between successive terms: a_{n+1}/a_n = r for all n.

1) Relevant concepts
- Geometric series: terms a_n = a r^{n} (or a r^{n-1}) so a_{n+1}/a_n = r constant.
- To test a candidate, compute the ratio a_{n+1}/a_n and check if it is independent of n.

2) Step-by-step reasoning (apply the test to each option)
- Option b: a_n = 3^{n+1}/2^n.
  Compute a_{n+1}/a_n = [3^{(n+2)}/2^{(n+1)}] ÷ [3^{(n+1)}/2^n]
  = (3^{n+2}/2^{n+1}) * (2^n/3^{n+1})
  = 3/2.
  This ratio is constant (independent of n), so option b is geometric. Also note a_n = 3*(3/2)^n, explicitly in geometric form with common ratio r = 3/2.

3) Intermediate calculations (shown above)
- For b: a_{n+1}/a_n = 3/2 (constant).

4) Why the correct answer is correct
- Because successive term ratio = 3/2 for every n, the sequence of terms is geometric and the series is a geometric series (though with |r|>1 it diverges, the question only asks to identify which is geometric).

5) Why the other options are incorrect (briefly)
- a: a_n = 1/n → a_{n+1}/a_n = n/(n+1), depends on n, not constant (harmonic series).
- c: a_n = (2n+1)/n^2 → a_{n+1}/a_n = [(2n+3)/(n+1)^2]·[n^2/(2n+1)], depends on n, not constant.
- d: a_n = n/2^n → a_{n+1}/a_n = ((n+1)/2^{n+1})/ (n/2^n) = ((n+1)/n)·(1/2), not constant.
- e: a_n = 1/n^2 → a_{n+1}/a_n = n^2/(n+1)^2, depends on n.

6) Conclusion
Option b (∑_{n=1}^∞ 3^{n+1}/2^n) is the geometric series because its consecutive-term ratio is the constant 3/2. This matches the provided answer.

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Which one of the following is a geometric series?单项选择题

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