Asked by Geesh
b(n) = 1 \left(-2\right)^{n - 1}
All Answers 1
Answered by
GPT-5 mini
AI
You can simplify it to
b(n) = (−2)^(n−1) = (−1)^(n−1) 2^(n−1).
Properties:
- First terms (n = 1,2,3,4,5): 1, −2, 4, −8, 16, ...
- Recurrence: b(1) = 1, b(n) = −2·b(n−1) for n ≥ 2.
- Sum of first n terms (geometric series): S_n = 1 + (−2) + (−2)^2 + … + (−2)^(n−1) = (1 − (−2)^n)/3.
- Because |−2| > 1, b(n) diverges in magnitude as n → ∞ (no finite limit).
b(n) = (−2)^(n−1) = (−1)^(n−1) 2^(n−1).
Properties:
- First terms (n = 1,2,3,4,5): 1, −2, 4, −8, 16, ...
- Recurrence: b(1) = 1, b(n) = −2·b(n−1) for n ≥ 2.
- Sum of first n terms (geometric series): S_n = 1 + (−2) + (−2)^2 + … + (−2)^(n−1) = (1 − (−2)^n)/3.
- Because |−2| > 1, b(n) diverges in magnitude as n → ∞ (no finite limit).
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