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COMEDK · Maths · 5. Sequences and Series

\(0.2+0.22+0.022+\ldots \ldots \ldots\) up to \(n\) terms is equal to

  1. A \(\dfrac{n}{9}\left(1-10^{-n}\right)\)
  2. B \(\dfrac{2}{9}-\dfrac{2}{81}\left(1-10^{-n}\right)\)
  3. C \(\dfrac{2}{9}\left(1-10^{-n}\right)\)
  4. D \(\dfrac{2}{9}\left[n-\dfrac{1}{9}\left(1-10^{-n}\right)\right]\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\dfrac{2}{9}\left[n-\dfrac{1}{9}\left(1-10^{-n}\right)\right]\)

Step-by-step Solution

Detailed explanation

The given series is \(S_n = 0.2 + 0.22 + 0.222 + \ldots\) up to \(n\) terms. We can write each term as \(T_k = 2 \times (0.1 + 0.01 + 0.001 + \ldots + 10^{-k}) = 2 \times \sum_{i=1}^{k} 10^{-i}\). Alternatively, express the terms as…