The sum of the series formed by the sequence `3,sqrt(3),1......`
upto infinity 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)`

Explanation

To find the sum of the series formed by the sequence \(3, \sqrt{3}, 1, \ldots\) up to infinity, we first need to identify the nature of the series.

### Step 1: Identify the first term and the common ratio
The first term \(a\) of the series is \(3\). The second term is \(\sqrt{3}\) and the third term is \(1\).

To find the common ratio \(r\), we can divide the second term by the first term:
\[ r = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

Now we can check the ratio between the second and third terms:
\[ r = \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]

Thus, the series is a geometric progression (GP) with:
- First term \(a = 3\)
- Common ratio \(r = \frac{1}{\sqrt{3}}\)

### Step 2: Check the condition for convergence
For a geometric series to converge, the absolute value of the common ratio must be less than 1:
\[ |r| < 1 \implies \left|\frac{1}{\sqrt{3}}\right| < 1 \]
This condition is satisfied since \(\sqrt{3} > 1\).

### Step 3: Use the formula for the sum of an infinite geometric series
The sum \(S\) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values of \(a\) and \(r\):
\[ S = \frac{3}{1 - \frac{1}{\sqrt{3}}} \]

### Step 4: Simplify the expression
Now, we simplify the denominator:
\[ 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}} \]
Thus, we can rewrite the sum as:
\[ S = \frac{3}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = 3 \cdot \frac{\sqrt{3}}{\sqrt{3} - 1} = \frac{3\sqrt{3}}{\sqrt{3} - 1} \]

### Step 5: Rationalize the denominator
To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{3} + 1\):
\[ S = \frac{3\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{3\sqrt{3}(\sqrt{3} + 1)}{2} \]
This simplifies to:
\[ S = \frac{3\sqrt{3}^2 + 3\sqrt{3}}{2} = \frac{9 + 3\sqrt{3}}{2} \]

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