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 identify the type of series we are dealing with.

### Step 1: Identify the series type
The series appears to be a geometric progression (GP). The first term \(a\) is \(1\), and the second term is \(\sqrt{3}\).

### Step 2: Find the common ratio \(r\)
To find the common ratio \(r\), we can divide the second term by the first term:
\[ r = \frac{a_2}{a_1} = \frac{\sqrt{3}}{1} = \sqrt{3} \]

### Step 3: Confirm the common ratio
We can also check the ratio between the third term and the second term:
\[ r = \frac{a_3}{a_2} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
This confirms that the common ratio is indeed \(\sqrt{3}\).

### Step 4: Use the formula for the sum of the first \(n\) terms of a GP
The formula for the sum \(S_n\) of the first \(n\) terms of a geometric progression is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Where:
- \(a\) is the first term,
- \(r\) is the common ratio,
- \(n\) is the number of terms.

### Step 5: Substitute the values into the formula
Here, \(a = 1\), \(r = \sqrt{3}\), and \(n = 10\):
\[ S_{10} = 1 \cdot \frac{(\sqrt{3})^{10} - 1}{\sqrt{3} - 1} \]

### Step 6: Simplify \((\sqrt{3})^{10}\)
Calculating \((\sqrt{3})^{10}\):
\[ (\sqrt{3})^{10} = 3^{5} = 243 \]

### Step 7: Substitute back into the sum formula
Now substituting this back into our equation:
\[ S_{10} = \frac{243 - 1}{\sqrt{3} - 1} = \frac{242}{\sqrt{3} - 1} \]

### Step 8: Rationalize the denominator
To simplify \(\frac{242}{\sqrt{3} - 1}\), we multiply the numerator and the denominator by the conjugate of the denominator, \(\sqrt{3} + 1\):
\[ S_{10} = \frac{242(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{242(\sqrt{3} + 1)}{3 - 1} = \frac{242(\sqrt{3} + 1)}{2} \]

### Step 9: Final simplification
Now, simplifying further:
\[ S_{10} = 121(\sqrt{3} + 1) \]

### Conclusion
Thus, the sum of the first 10 terms of the series is:
\[ \boxed{121(\sqrt{3} + 1)} \]