Question

$0.2 + 0.22 + 0.222 + \ldots$ to n terms is equal to

• A. $\frac{2}{9} - \frac{2}{81}(1-10^{-n})$
• B. $\frac{n}{9} - \frac{1}{9}(1-10^{-n})$
• C. $\frac{2}{9} - \frac{1}{9}(1-10^{-n})$
• D. $\frac{2}{9}$

Hint:

$0.\overline{2} = 2/9$, $0.22 = 2/9 - 2/90$, etc.

Answer 1

The Correct Option is A

To solve the given series and find the sum of the first \(n\) terms: \(0.2 + 0.22 + 0.222 + \ldots\), we need to recognize a pattern or formula.

1. Each term in the series can be expressed in the form: \(0.2, 0.22, 0.222,\ldots\).
2. Let's write down the decimal numbers in fraction form:
   • \(0.2 = \frac{2}{10}\)
   • \(0.22 = \frac{22}{100} = \frac{2 \times (10+2)}{100}\)
   • \(0.222 = \frac{222}{1000} = \frac{2 \times (100+20+2)}{1000}\)
3. The general \(n^{th}\) term can be represented as: \(\frac{2 \times (10^{n-1} + 10^{n-2} + \ldots + 1)}{10^n}\).
4. The sum of each part of the sequence is a geometric progression with the first term \(a = 1\) and common ratio \(r = 10\). Therefore, the sum for \(n\) terms is: \(1 + 10 + 10^2 + \ldots + 10^{n-1} = \frac{10^n - 1}{10 - 1} = \frac{10^n - 1}{9}\).
5. Thus, the sum of the series up to \(n\) terms becomes: \(\text{Sum} = \sum_{k=1}^n \frac{2 \times (10^k - 1)}{10^k}\).
6. Simplifying the above summation, we get the sum: \(\text{Sum} = 2 \left( \frac{10^n - 1}{9 \times 10^n} + \frac{1}{10} + \frac{1}{10^2} + \ldots + \frac{1}{10^{n-1}} \right) = 2 \left( \frac{1}{9} - \frac{1}{9 \times 10^n} + \frac{1}{9} - \frac{1}{9 \times 10^n} \right)\).
7. Simplifying further: \(\text{Sum} = \frac{2}{9} - \frac{2}{81}(1 - 10^{-n})\).

Comparing our result with the given options, we confirm that the correct option is: \(\frac{2}{9} - \frac{2}{81}(1-10^{-n})\).