What is the sum of the series 0.5 + 0.55 + 0.555 + … to n terms?
Finding the Sum of the Given Series
Let's find the sum of the series \(S_n = 0.5 + 0.55 + 0.555 + \dots\) up to \(n\) terms. This is a common type of series problem that can be solved by transforming the terms into a more manageable form involving a geometric progression.
We can write each term by factoring out 5:
\(S_n = 5 \times 0.1 + 5 \times 0.11 + 5 \times 0.111 + \dots\) up to \(n\) terms.
\(S_n = 5 (0.1 + 0.11 + 0.111 + \dots)\) up to \(n\) terms.
Now, let's multiply and divide by 9 inside the bracket. This step is crucial as it helps us transform the decimal terms into a difference involving powers of 10.
\(S_n = \frac{5}{9} (9 \times 0.1 + 9 \times 0.11 + 9 \times 0.111 + \dots)\) up to \(n\) terms.
\(S_n = \frac{5}{9} (0.9 + 0.99 + 0.999 + \dots)\) up to \(n\) terms.
We can rewrite each term inside the bracket using the form \(1 - \frac{1}{10^k}\):
- \(0.9 = 1 - 0.1 = 1 - \frac{1}{10^1}\)
- \(0.99 = 1 - 0.01 = 1 - \frac{1}{10^2}\)
- \(0.999 = 1 - 0.001 = 1 - \frac{1}{10^3}\)
- ... and so on, up to the n-th term \(1 - \frac{1}{10^n}\)
Substituting these forms into the sum:
\(S_n = \frac{5}{9} \left[ \left(1 - \frac{1}{10}\right) + \left(1 - \frac{1}{10^2}\right) + \left(1 - \frac{1}{10^3}\right) + \dots + \left(1 - \frac{1}{10^n}\right) \right]\)
Now, we can rearrange the terms by grouping the '1's together and the fractional terms together:
\(S_n = \frac{5}{9} \left[ (1 + 1 + 1 + \dots \text{n terms}) - \left(\frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \dots + \frac{1}{10^n}\right) \right]\)
The sum of '1' repeated \(n\) times is simply \(n\).
The terms in the second bracket form 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 \(n\) terms is given by the formula \(S_{GP} = a \frac{(1 - r^n)}{1 - r}\) (when \(r \neq 1\)).
Substituting the values for this specific geometric series:
\(S_{GP} = \frac{1}{10} \frac{\left(1 - \left(\frac{1}{10}\right)^n\right)}{1 - \frac{1}{10}}\)
\(S_{GP} = \frac{1}{10} \frac{\left(1 - \frac{1}{10^n}\right)}{\frac{9}{10}}\)
\(S_{GP} = \frac{1}{10} \times \frac{10}{9} \left(1 - \frac{1}{10^n}\right)\)
\(S_{GP} = \frac{1}{9} \left(1 - \frac{1}{10^n}\right)\)
Now, substitute this sum of the geometric series back into the expression for \(S_n\):
\(S_n = \frac{5}{9} \left[ n - \frac{1}{9} \left(1 - \frac{1}{10^n}\right) \right]\)
This is the formula for the sum of the given series to \(n\) terms.
Comparing with Options
Let's look at the derived formula and compare it with the provided options:
- Option 1: \(\frac{5}{9}\left[ {n - \frac{2}{9}\left( {1 - \frac{1}{{{{10}^n}}}} \right)} \right]\)
- Option 2: \(\frac{1}{9}\left[ {5 - \frac{2}{9}\left( {1 - \frac{1}{{{{10}^n}}}} \right)} \right]\)
- Option 3: \(\frac{1}{9}\left[ {n - \frac{5}{9}\left( {1 - \frac{1}{{{{10}^n}}}} \right)} \right]\)
- Option 4: \(\frac{5}{9}\left[ {n - \frac{1}{9}\left( {1 - \frac{1}{{{{10}^n}}}} \right)} \right]\)
Our derived formula is \(\frac{5}{9} \left[ n - \frac{1}{9} \left(1 - \frac{1}{10^n}\right) \right]\), which exactly matches Option 4.
| Step | Calculation |
|---|---|
| 1 | Write the sum: \(S_n = 0.5 + 0.55 + \dots\) |
| 2 | Factor out 5: \(S_n = 5(0.1 + 0.11 + \dots)\) |
| 3 | Multiply/Divide by 9: \(S_n = \frac{5}{9}(0.9 + 0.99 + \dots)\) |
| 4 | Rewrite terms: \(S_n = \frac{5}{9}((1-0.1) + (1-0.01) + \dots)\) |
| 5 | Separate sums: \(S_n = \frac{5}{9}[n - (0.1 + 0.01 + \dots)]\) |
| 6 | Sum GP: \(0.1 + 0.01 + \dots = \frac{1}{9}(1 - \frac{1}{10^n})\) |
| 7 | Substitute GP sum: \(S_n = \frac{5}{9}[n - \frac{1}{9}(1 - \frac{1}{10^n})] \) |
Revision Table: Key Concepts for Series Sum
| Concept | Description | Formula Example |
|---|---|---|
| Arithmetic Progression (AP) | A sequence where the difference between consecutive terms is constant. | Sum of first \(n\) terms: \(S_n = \frac{n}{2}(2a + (n-1)d)\) |
| Geometric Progression (GP) | A sequence where the ratio between consecutive terms is constant. | Sum of first \(n\) terms: \(S_n = a\frac{(r^n - 1)}{r - 1}\) or \(a\frac{(1 - r^n)}{1 - r}\) (\(r \neq 1\)) |
| Sum to Infinity (GP) | Sum of a GP when the number of terms is infinite and \(|r| < 1\). | \(S_\infty = \frac{a}{1 - r}\) |
| Manipulating Decimal Series | Technique involves factoring and rewriting terms (e.g., using 1-powers of 10) to convert to standard series forms (like GP). | Example: \(0.333\dots = 3 \times 0.111\dots = 3 \times \frac{1}{9} (0.999\dots) = \frac{1}{3} (1 - 0.00\dots1)\) or directly as a GP. |
Additional Information on Series Summation
Understanding how to sum different types of series is a fundamental skill in mathematics. The series \(0.5 + 0.55 + 0.555 + \dots\) is an example of a series that isn't immediately an arithmetic or geometric progression, but can be transformed into one (or a combination of basic series) through algebraic manipulation.
The key technique used here was to convert the repeating decimal structure into a form that relates to powers of 10. By factoring out 5 and then introducing division by 9 (and multiplication by 9), we could rewrite terms like 0.555 as \(5 \times 0.111 = 5 \times \frac{1}{9} \times 0.999 = \frac{5}{9} (1 - 0.001)\). This breaks down each term into a simple part (the '1') and a geometric part (\(\frac{1}{10^k}\)). Summing these parts separately simplifies the problem significantly.
This method is generally applicable to series of the form \(a.aa\dots a + a.aaa\dots a + \dots\), where 'a' is a single digit.
For example, to find the sum of \(0.7 + 0.77 + 0.777 + \dots\) to \(n\) terms, you would start by factoring out 7, then multiplying and dividing by 9, and proceed similarly:
\(S_n = 7(0.1 + 0.11 + 0.111 + \dots)\)
\(S_n = \frac{7}{9}(0.9 + 0.99 + 0.999 + \dots)\)
\(S_n = \frac{7}{9}[(1-0.1) + (1-0.01) + (1-0.001) + \dots]\)
\(S_n = \frac{7}{9}[n - (0.1 + 0.01 + 0.001 + \dots)]\)
\(S_n = \frac{7}{9}[n - \frac{1}{9}(1 - \frac{1}{10^n})]\)
This shows the general pattern for such series.
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