Question:medium

The series \(\displaystyle\sum_{n=0}^{\infty}(2x)^n\) converges if ____.

Show Hint

For a geometric series \(\sum r^n\), convergence occurs only when \(|r|<1\). Always check the end points separately.
  • \(-1\leq x\leq 1\)
  • \(-\dfrac{1}{2}<x<\dfrac{1}{2}\)
  • \(-2<x<2\)
  • \(-\dfrac{1}{2}\leq x\leq\dfrac{1}{2}\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The given series is a geometric series. A geometric series has the form \(\sum_{n=0}^\infty ar^n\). We need to find the values of \(x\) for which this series converges.

Step 2: Key Formula or Approach:

A geometric series \(\sum_{n=0}^\infty ar^n\) converges if and only if the absolute value of the common ratio \(r\) is less than 1, i.e., \(|r| < 1\). The series diverges if \(|r| \ge 1\). For the given series, we need to identify the common ratio \(r\) and apply this condition.

Step 3: Detailed Explanation:

The given series is \(\sum_{n=0}^\infty (2x)^n\). This is a geometric series with the first term \(a = (2x)^0 = 1\) and the common ratio \(r = 2x\). For the series to converge, we must have \(|r| < 1\). \[ |2x| < 1 \] This inequality can be split into two parts: \[ -1 < 2x < 1 \] To solve for \(x\), we divide all parts of the inequality by 2: \[ \frac{-1}{2} < x < \frac{1}{2} \] This is the interval of convergence for the series.

Step 4: Final Answer:

The series converges if \(-\frac{1}{2} < x < \frac{1}{2}\), which corresponds to option (B).
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