The sum of 24 terms of the following series `sqrt(2)+sqrt(8)+sqrt(18)+sqrt(32)+ . .. ` is

Options

A. 300
B. `300sqrt(2)`
C. `200sqrt(2)`
D. none of these

Answer

To find the sum of the first 24 terms of the series \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \ldots \), we will follow these steps:

### Step 1: Identify the terms of the series
The given series consists of terms that can be expressed in terms of \( \sqrt{2} \):
- \( \sqrt{2} = \sqrt{2} \)
- \( \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} \)
- \( \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \)
- \( \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \)

From this, we can see that the terms can be rewritten as:
- \( \sqrt{2} \)
- \( 2\sqrt{2} \)
- \( 3\sqrt{2} \)
- \( 4\sqrt{2} \)
- ...
- \( n\sqrt{2} \) where \( n \) is the term number.

### Step 2: General term of the series
The \( n \)-th term of the series can be expressed as:
\[ T_n = n\sqrt{2} \]

### Step 3: Find the sum of the first 24 terms
The sum of the first \( n \) terms of the series can be calculated using the formula:
\[ S_n = T_1 + T_2 + T_3 + \ldots + T_n = \sqrt{2}(1 + 2 + 3 + \ldots + n) \]

The sum of the first \( n \) natural numbers is given by:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]

For \( n = 24 \):
\[ \text{Sum} = \frac{24(24 + 1)}{2} = \frac{24 \times 25}{2} = \frac{600}{2} = 300 \]

### Step 4: Multiply by \( \sqrt{2} \)
Now, substituting back into the sum formula:
\[ S_{24} = \sqrt{2} \times 300 = 300\sqrt{2} \]

### Final Answer
The sum of the first 24 terms of the series is:
\[ \boxed{300\sqrt{2}} \]