Question:

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

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Use derivative of GP when n appears in numerator.
Updated On: Apr 15, 2026
  • 200/891
  • 2000/9801
  • 1000/9801
  • None of these
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The Correct Option is B

Solution and Explanation

Concept: Write terms as: \[ 0.2 = \frac{2}{10},\quad 0.004=\frac{4}{10^3},\dots \] \[ = \sum \frac{2n}{10^{2n-1}} \]

Step 1:
Transform series.
Convert into derivative of GP.

Step 2:
Use known result.
\[ \sum_{n=1}^{\infty} \frac{n}{r^n} = \frac{r}{(r-1)^2} \]

Step 3:
Final.
\[ = \frac{2000}{9801} \]
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