Question

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

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

Hint:

For decimal series, convert into fractions and identify a pattern that allows splitting into geometric progressions.

Answer 1

The Correct Option is B

Step 1: Understand the pattern of the series.
The given series is:
\[ 0.2 + 0.22 + 0.022 + \cdots \]
Convert each term into fractional form:
\[ 0.2 = \frac{2}{10}, \quad 0.22 = \frac{22}{100}, \quad 0.022 = \frac{22}{1000}, \dots \]

Step 2: Express each term systematically.
We can write each term as:
\[ T_1 = \frac{2}{10}, \quad T_2 = \frac{22}{100}, \quad T_3 = \frac{22}{1000}, \dots \]
Notice that after the first term, numerator remains 22 and denominator increases as powers of 10.

Step 3: Split the series into two parts.
We rewrite each term as:
\[ 0.2 = \frac{2}{10}, \quad 0.22 = \frac{2}{10} + \frac{2}{100}, \quad 0.022 = \frac{2}{100} + \frac{2}{1000}. \]
So the sum becomes:
\[ S = \sum_{k=1}^{n} \left(\frac{2}{10^k} + \frac{2}{10^{k+1}}\right). \]

Step 4: Rearrange the summation.
\[ S = 2\sum_{k=1}^{n} \frac{1}{10^k} + 2\sum_{k=2}^{n+1} \frac{1}{10^k}. \]
Now combine both sums:
\[ S = 2\left[\frac{1}{10} + \frac{1}{10^2} + \cdots + \frac{1}{10^n}\right] + 2\left[\frac{1}{10^2} + \cdots + \frac{1}{10^{n+1}}\right]. \]

Step 5: Use GP sum formula.
The sum of GP is:
\[ \sum_{k=1}^{n} \frac{1}{10^k} = \frac{1}{9}(1-10^{-n}). \]
Similarly,
\[ \sum_{k=2}^{n+1} \frac{1}{10^k} = \frac{1}{9}(1-10^{-n}). \]

Step 6: Substitute values.
\[ S = 2\left[\frac{1}{9}(1-10^{-n})\right] + 2\left[\frac{1}{9}(1-10^{-n}) - \frac{1}{10}\right]. \]
Simplifying further:
\[ S = \frac{2}{9}(1-10^{-n}) + \frac{2}{9}(1-10^{-n}) - \frac{2}{10}. \]

Step 7: Final simplification.
\[ S = \frac{2}{9}\left[n - \frac{1}{9}(1-10^{-n})\right]. \]
Final Answer:
\[ \boxed{\frac{2}{9}\left[n - \frac{1}{9}(1-10^{-n})\right]}. \]