Find the sum of 10 terms of the series `1+sqrt(3)+3+….`

Explanation

To find the sum of the first 10 terms of the series \(1 + \sqrt{3} + 3 + \ldots\), we first need to determine if this series is a geometric progression (GP).

### Step 1: Identify the terms and check for GP
The first three terms of the series are:
- \(a_1 = 1\)
- \(a_2 = \sqrt{3}\)
- \(a_3 = 3\)

To check if this is a GP, we need to find the common ratio \(r\) between the terms:
\[ r = \frac{a_2}{a_1} = \frac{\sqrt{3}}{1} = \sqrt{3} \]
\[ r = \frac{a_3}{a_2} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

Since the common ratio is the same, the series is indeed a geometric progression.

### Step 2: Identify the first term and common ratio
From the above calculations, we have:
- First term \(a = 1\)
- Common ratio \(r = \sqrt{3}\)

### Step 3: Use the formula for the sum of the first \(n\) terms of a GP
The formula for the sum of the first \(n\) terms of a GP is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
where \(n\) is the number of terms.

### Step 4: Substitute values into the formula
We want to find the sum of the first 10 terms (\(n = 10\)):
\[ S_{10} = \frac{1 \cdot ((\sqrt{3})^{10} - 1)}{\sqrt{3} - 1} \]

### Step 5: Calculate \((\sqrt{3})^{10}\)
\[ (\sqrt{3})^{10} = (3^{1/2})^{10} = 3^{5} = 243 \]

### Step 6: Substitute back into the sum formula
Now substituting this value back into the sum formula:
\[ S_{10} = \frac{243 - 1}{\sqrt{3} - 1} = \frac{242}{\sqrt{3} - 1} \]

### Step 7: Rationalize the denominator
To simplify \(\frac{242}{\sqrt{3} - 1}\), we multiply the numerator and denominator by the conjugate of the denominator:
\[ S_{10} = \frac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{242(\sqrt{3} + 1)}{3 - 1} = \frac{242(\sqrt{3} + 1)}{2} \]

### Step 8: Final result
Thus, the sum of the first 10 terms is:
\[ S_{10} = 121(\sqrt{3} + 1) \]