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.
Similar Questions
If g is the geometric mean of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, then which one of the following is correct?
What is the greatest value of the positive integer n satisfying the condition \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}} < 2 - \frac{1}{{1000}}?\)
A geometric progression (GP) consists of 200 terms. If the sum of odd terms of the GP is m, and the sum of even terms of the GP is n, then what is its common ratio?
If the second term of a GP is 2 and the sum of its infinite terms is 8, then the GP is
If m is the geometric mean of \({\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right)^{\log \left( {{\rm{yz}}} \right)}},{\rm{\;}}{\left( {\frac{{\rm{z}}}{{\rm{x}}}} \right)^{\log \left( {{\rm{zx}}} \right)}}{\rm{\;and\;}}{\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)^{\log \left( {{\rm{xy}}} \right)}}\) then what is the value of m?
The value of the infinite product \({6^{\frac{1}{2}}} \times {6^{\frac{1}{2}}} \times {6^{\frac{3}{8}}} \times {6^{\frac{1}{4}}} \times \ldots \) is
The geometric mean of the observations x 1, x 2, x 3, … x nis G 1. The geometric mean of the observations y 1, y 2, y 3,… y nis G 2. The geometric mean of observations \(\frac{{{{\rm{x}}_1}}}{{{{\rm{y}}_1}}},\frac{{{{\rm{x}}_2}}}{{{{\rm{y}}_2}}},\frac{{{{\rm{x}}_3}}}{{{{\rm{y}}_3}}}, \ldots \frac{{{{\rm{x}}_{\rm{n}}}}}{{{{\rm{y}}_{\rm{n}}}}}\) is
If p, q, r are in one geometric progression and a, b, c are in another geometric progression, then ap, bq, cr are in
Let t1, t2, t3 ... be in GP. What is \(\rm \left(t_1 t_3 \ldots t_{21}\right)^{\frac{1}{11}}\) equal to ?
Consider the following statements:
1. If each term of a GP is multiplied by same non-zero number, then the resulting sequence is also a GP.
2. If each term of a GP is divided by same non-zero number, then the resulting sequence is also a GP.
Which of the above statements is/are correct?
Important Questions from Geometric Progressions
If g is the geometric mean of 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, then which one of the following is correct?
What is the greatest value of the positive integer n satisfying the condition \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{{{2^{{\rm{n}} - 1}}}} < 2 - \frac{1}{{1000}}?\)
The sum of even numbers from 1 to 40 is:
The minimum value of the sum of real numbers a-5, a-4, 3a-3, 1, a8 and a10 with a > 0 is:
The arithmetic mean, geometric mean and median of six positive numbers a, a, b, b, c, c where a < b < c are \(\frac 7 3,\) 2, 2 respectively. Then what is the sum of the squares of all the six numbers?
