The sum of the series `1+ 2/3+ (1)/(3 ^(2)) + (2 )/(3 ^(3)) + (1)/(3 ^(4)) + (2)/(3 ^(5)) + (1)/(3 ^(6))+ (2)/(3 ^(7))+ ……` upto infinite terms is equal to :

Options

A. `15/8`
B. `8/15`
C. `27/8`
D. `21/8`

Answer

To find the sum of the series

\[ S = 1 + \frac{2}{3} + \frac{1}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} + \frac{2}{3^5} + \frac{1}{3^6} + \frac{2}{3^7} + \ldots \]

we can group the terms based on their numerators:

1. **Group the series**:
- The series can be separated into two parts:
\[ S = (1 + \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \ldots) + (\frac{2}{3} + \frac{2}{3^3} + \frac{2}{3^5} + \frac{2}{3^7} + \ldots) \]

2. **Identify the first series (S1)**:
- The first series is:
\[ S_1 = 1 + \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \ldots \]
- This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = \frac{1}{3^2} = \frac{1}{9} \).

3. **Sum of the first series (S1)**:
- The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
- Therefore, for \( S_1 \):
\[ S_1 = \frac{1}{1 - \frac{1}{9}} = \frac{1}{\frac{8}{9}} = \frac{9}{8} \]

4. **Identify the second series (S2)**:
- The second series is:
\[ S_2 = \frac{2}{3} + \frac{2}{3^3} + \frac{2}{3^5} + \frac{2}{3^7} + \ldots \]
- This can be factored as:
\[ S_2 = 2 \left( \frac{1}{3} + \frac{1}{3^3} + \frac{1}{3^5} + \ldots \right) \]
- The series inside the parentheses is also a geometric series with first term \( a = \frac{1}{3} \) and common ratio \( r = \frac{1}{9} \).

5. **Sum of the second series (S2)**:
- Therefore, for \( S_2 \):
\[ S_2 = 2 \cdot \frac{\frac{1}{3}}{1 - \frac{1}{9}} = 2 \cdot \frac{\frac{1}{3}}{\frac{8}{9}} = 2 \cdot \frac{3}{8} = \frac{6}{8} = \frac{3}{4} \]

6. **Combine the sums**:
- Now we can combine \( S_1 \) and \( S_2 \):
\[ S = S_1 + S_2 = \frac{9}{8} + \frac{3}{4} \]
- To add these fractions, we need a common denominator. The least common multiple of 8 and 4 is 8:
\[ \frac{3}{4} = \frac{6}{8} \]
- Thus,
\[ S = \frac{9}{8} + \frac{6}{8} = \frac{15}{8} \]

Therefore, the sum of the series is

\[ \boxed{\frac{15}{8}} \]