The sum of the series formed by the sequence 3, `sqrt3`, 1, . . . Upto infinite is

Options

A. `(3sqrt(3)(sqrt(3)+1))/(2)`
B. `(3sqrt(3)(sqrt(3)-1))/(2)`
C. `(3(sqrt(3)+1))/(2)`
D. `(3(sqrt(3)-1))/(2)`

Answer

To find the sum of the series formed by the sequence 3, √3, 1, ..., up to infinity, we can recognize that this series is a geometric progression (GP).

### Step-by-step Solution:

1. **Identify the first term (a)**:
The first term of the series is \( a = 3 \).

**Hint**: The first term is simply the first number in the sequence.

2. **Identify the common ratio (r)**:
The common ratio \( r \) can be found by dividing the second term by the first term:
\[ r = \frac{\sqrt{3}}{3} \]
We can simplify this:
\[ r = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

**Hint**: The common ratio is found by taking the ratio of consecutive terms in the series.

3. **Check if the series converges**:
For a geometric series to converge, the absolute value of the common ratio must be less than 1:
\[ |r| < 1 \implies \left|\frac{1}{\sqrt{3}}\right| < 1 \]
Since \( \sqrt{3} > 1 \), this condition is satisfied.

**Hint**: A geometric series converges if the common ratio is between -1 and 1.

4. **Use the formula for the sum of an infinite geometric series**:
The sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values of \( a \) and \( r \):
\[ S = \frac{3}{1 - \frac{1}{\sqrt{3}}} \]

**Hint**: The formula for the sum of a GP is essential for finding the total sum.

5. **Simplify the denominator**:
To simplify \( 1 - \frac{1}{\sqrt{3}} \), find a common denominator:
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]

**Hint**: Finding a common denominator helps in simplifying fractions.

6. **Substitute back into the sum formula**:
Now substitute this back into the sum formula:
\[ S = \frac{3}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = 3 \cdot \frac{\sqrt{3}}{\sqrt{3} - 1} = \frac{3\sqrt{3}}{\sqrt{3} - 1} \]

**Hint**: Multiplying by the reciprocal of the denominator can help simplify the expression.

7. **Rationalize the denominator**:
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{3} + 1 \):
\[ S = \frac{3\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{3\sqrt{3}(\sqrt{3} + 1)}{2} \]

**Hint**: Rationalizing the denominator can make the expression easier to work with.

8. **Final simplification**:
Expanding the numerator:
\[ S = \frac{3\sqrt{3} \cdot \sqrt{3} + 3\sqrt{3}}{2} = \frac{9 + 3\sqrt{3}}{2} \]

**Hint**: Expanding helps in simplifying the expression further.

### Final Answer:
The sum of the series is:
\[ S = \frac{9 + 3\sqrt{3}}{2} \]