Understanding the Series Sum Problem

The problem asks for the sum of a specific type of series: 0.3 + 0.33 + 0.333 + … up to 'n' terms. This is not a standard arithmetic or geometric progression directly, but we can manipulate it to use standard series summation techniques, particularly those related to geometric series.

Breaking Down the Series Terms

Let the sum of the series up to 'n' terms be \( S_n \). The terms can be written as fractions:

• First term: \( 0.3 = \frac{3}{10} \)
• Second term: \( 0.33 = \frac{33}{100} \)
• Third term: \( 0.333 = \frac{333}{1000} \)
• The \( k \)-th term: \( \frac{33\dots3}{10^k} \) (with \( k \) digits of 3 in the numerator)

So the series is:

\[ S_n = \frac{3}{10} + \frac{33}{100} + \frac{333}{1000} + \dots + \text{n terms} \]

Manipulating the Terms for Easier Summation

We can factor out 3 from each term:

\[ S_n = 3 \left( \frac{1}{10} + \frac{11}{100} + \frac{111}{1000} + \dots + \text{n terms} \right) \]

Now, let's look at the terms inside the parenthesis: \( \frac{1}{10}, \frac{11}{100}, \frac{111}{1000}, \dots \). These can be rewritten by multiplying and dividing by 9:

• \( \frac{1}{10} = \frac{1}{9} \times \frac{9}{10} \)
• \( \frac{11}{100} = \frac{1}{9} \times \frac{99}{100} \)
• \( \frac{111}{1000} = \frac{1}{9} \times \frac{999}{1000} \)
• The \( k \)-th term: \( \frac{1}{9} \times \frac{99\dots9}{10^k} \) (with \( k \) digits of 9 in the numerator)

Note that the numerator \( 99\dots9 \) (with \( k \) digits) is equal to \( 10^k - 1 \). So, the \( k \)-th term can be written as \( \frac{1}{9} \times \frac{10^k - 1}{10^k} = \frac{1}{9} \left( 1 - \frac{1}{10^k} \right) \).

Rewriting the Series using the New Form

Substituting this back into the expression for \( S_n \):

\[ S_n = 3 \left[ \frac{1}{9}\left(1 - \frac{1}{10}\right) + \frac{1}{9}\left(1 - \frac{1}{10^2}\right) + \frac{1}{9}\left(1 - \frac{1}{10^3}\right) + \dots + \frac{1}{9}\left(1 - \frac{1}{10^n}\right) \right] \]

Factor out \( \frac{1}{9} \) from the bracket:

\[ S_n = 3 \times \frac{1}{9} \left[ \left(1 - \frac{1}{10}\right) + \left(1 - \frac{1}{10^2}\right) + \left(1 - \frac{1}{10^3}\right) + \dots + \left(1 - \frac{1}{10^n}\right) \right] \]

\[ S_n = \frac{1}{3} \left[ (1+1+1+\dots+\text{n times}) - \left(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \dots + \frac{1}{10^n}\right) \right] \]

Summing the Individual Parts

The sum of 'n' terms of 1 is simply \( n \).

The second part \( \left(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \dots + \frac{1}{10^n}\right) \) is a geometric series with:

• First term \( a = \frac{1}{10} \)
• Common ratio \( r = \frac{1/10^2}{1/10} = \frac{1}{10} \)
• Number of terms = \( n \)

The sum of a geometric series with first term \( a \), common ratio \( r \), and \( n \) terms is given by \( S_{n, GP} = a \frac{1 - r^n}{1 - r} \) (when \( r \neq 1 \)).

Using this formula for the geometric series part:

\[ \text{Sum of GP} = \frac{1}{10} \times \frac{1 - (\frac{1}{10})^n}{1 - \frac{1}{10}} \]

\[ \text{Sum of GP} = \frac{1}{10} \times \frac{1 - \frac{1}{10^n}}{\frac{9}{10}} \]

\[ \text{Sum of GP} = \frac{1}{10} \times \frac{10}{9} \times \left(1 - \frac{1}{10^n}\right) \]

\[ \text{Sum of GP} = \frac{1}{9} \left(1 - \frac{1}{10^n}\right) \]

Combining the Sums

Now substitute the sums of the two parts back into the expression for \( S_n \):

\[ S_n = \frac{1}{3} \left[ n - \frac{1}{9} \left(1 - \frac{1}{10^n}\right) \right] \]

This matches one of the given options.

Final Answer Derivation Summary

The sum of the series \( 0.3 + 0.33 + 0.333 + \dots \) up to \( n \) terms is found by:

1. Writing each term as a fraction.
2. Factoring out 3.
3. Rewriting the remaining terms using the form \( \frac{10^k - 1}{10^k} \).
4. Separating the series into a sum of 1s and a geometric series.
5. Summing the two parts.
6. Simplifying the result to get the final formula.

The resulting sum of the series formula is \( \frac{1}{3}\left[ {{\rm{n}} - \frac{1}{9}\left( {1 - \frac{1}{{{{10}^{\rm{n}}}}}} \right)} \right] \).

Revision Table: Key Concepts for Series Summation

Additional Information on Recurring Decimals and Series

The terms in the series (0.3, 0.33, 0.333, ...) are related to recurring decimals. For example, 0.3 is a terminating decimal. 0.333... is the recurring decimal \( 0.\bar{3} \). We know that \( 0.\bar{3} = \frac{3}{9} = \frac{1}{3} \). The given series approaches a value related to this as \( n \to \infty \).

As \( n \to \infty \), the term \( \frac{1}{10^n} \to 0 \). So the sum \( S_n \) approaches:

\[ \lim_{n \to \infty} S_n = \frac{1}{3} \left[ n - \frac{1}{9} \left(1 - 0\right) \right] = \frac{1}{3} \left[ n - \frac{1}{9} \right] \]

However, this limit involves 'n' and tends to infinity, which makes sense as we are adding increasingly larger terms (although the *difference* between consecutive terms decreases). The question asks for the sum up to 'n' terms, which is a finite sum.

Understanding how to convert repeating decimals or series involving repeated digits into fractions (like \( 0.\bar{a} = a/9 \), \( 0.aa\bar{a} = (aa)/99 \)) is helpful in manipulating such series problems.