Question:

The interval of convergence of the power series \[ \sum_{n=1}^{\infty} \frac{1}{(-3)^n + 2} \frac{(4x - 12)^n}{n^2 + 1} \] is

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When finding the interval of convergence of a power series, apply the ratio test to find the boundaries and check for absolute convergence.
Updated On: Nov 20, 2025
  • \( \frac{10}{4} \leq x \leq \frac{14}{4} \)
  • \( \frac{9}{4} \leq x \leq \frac{15}{4} \)
  • \( \frac{10}{4} \leq x \leq \frac{14}{4} \)
  • \( \frac{9}{4} \leq x \leq \frac{15}{4} \)
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The Correct Option is B

Solution and Explanation

Step 1: Apply the Ratio Test.
The ratio test can be applied to determine the interval of convergence. We examine the limit of the ratio of consecutive terms as \( n \to \infty \). The general term of the series is: \[ a_n = \frac{(4x - 12)^n}{(-3)^n + 2} \cdot \frac{1}{n^2 + 1}. \]
Step 2: Apply the ratio test to the terms.
We take the ratio of \( a_{n+1} \) to \( a_n \) and compute the limit: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \] This simplifies to a condition on \( x \), leading to the interval \( \frac{9}{4} \leq x \leq \frac{15}{4} \).
Step 3: Conclusion.
Thus, the correct answer is (B).
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