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

Answer

To find the sum of the first 10 terms of the series \(1 + \sqrt{3} + 3 + \ldots\), we first need to identify the nature of the series.

### Step 1: Identify the terms
The first few terms are:
- \(a_1 = 1\)
- \(a_2 = \sqrt{3}\)
- \(a_3 = 3\)

### Step 2: Determine the common ratio
To check if this is a geometric progression (GP), we need to find the common ratio \(r\):
- \(r = \frac{a_2}{a_1} = \frac{\sqrt{3}}{1} = \sqrt{3}\)
- \(r = \frac{a_3}{a_2} = \frac{3}{\sqrt{3}} = \sqrt{3}\)

Since both ratios are equal, the series is indeed a geometric progression with:
- 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 geometric progression when \(r > 1\) is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Here, we need to find \(S_{10}\):
- \(n = 10\)
- \(a = 1\)
- \(r = \sqrt{3}\)

### Step 4: Substitute the values into the formula
Substituting the values into the formula:
\[ S_{10} = 1 \cdot \frac{(\sqrt{3})^{10} - 1}{\sqrt{3} - 1} \]

### Step 5: Simplify the expression
Calculating \((\sqrt{3})^{10}\):
\[ (\sqrt{3})^{10} = (3^{1/2})^{10} = 3^{5} = 243 \]
Now, substituting back:
\[ S_{10} = \frac{243 - 1}{\sqrt{3} - 1} = \frac{242}{\sqrt{3} - 1} \]

### Step 6: Rationalize the denominator
To rationalize the denominator:
\[ S_{10} = \frac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{242(\sqrt{3} + 1)}{3 - 1} = \frac{242(\sqrt{3} + 1)}{2} \]
Thus, we have:
\[ S_{10} = 121(\sqrt{3} + 1) \]

### Final Answer
The sum of the first 10 terms of the series \(1 + \sqrt{3} + 3 + \ldots\) is:
\[ S_{10} = 121(\sqrt{3} + 1) \]