Question:medium

The sum of the series \(0.2 + 0.004 + 0.00006 + 0.0000008 + \dots\) is

Show Hint

Use derivative of GP when n appears in numerator.
Updated On: Apr 18, 2026
  • 200/891
  • 2000/9801
  • 1000/9801
  • None of these
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The given series can be expressed in terms of powers of 10. By observing the terms, we can identify it as an Arithmetic-Geometric Series (AGS) or use the sum formula for related power series.
: Key Formula or Approach:
The general term \( T_n \) is \( 2n \cdot 10^{-(2n-1)} \).
The sum of a power series \( \sum_{n=1}^{\infty} nx^{n} = \frac{x}{(1-x)^2} \) for \( |x|<1 \).
Step 2: Detailed Explanation:
Write the series in summation form: \[ S = \frac{2}{10^1} + \frac{4}{10^3} + \frac{6}{10^5} + \frac{8}{10^7} + \dots \] \[ S = \sum_{n=1}^{\infty} \frac{2n}{10^{2n-1}} \] Rearrange the terms to fit the standard power series form: \[ S = 2 \cdot 10^1 \sum_{n=1}^{\infty} \frac{n}{10^{2n}} = 20 \sum_{n=1}^{\infty} n \left( \frac{1}{100} \right)^n \] Let \( x = \frac{1}{100} = 0.01 \). \[ S = 20 \left[ \frac{x}{(1-x)^2} \right] \] Substitute \( x = 0.01 \): \[ S = 20 \left[ \frac{0.01}{(1 - 0.01)^2} \right] \] \[ S = \frac{0.2}{(0.99)^2} = \frac{0.2}{0.9801} \] \[ S = \frac{2000}{9801} \].
Step 3: Final Answer:
The sum of the series is \( 2000/9801 \).
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