What is the sum of n terms of the series `sqrt(2)+sqrt(8)+sqrt(18)+sqrt(32)+`. . . ?

Options

A. `(n(n-1))/(sqrt2)`
B. `sqrt(2)n(n+1)`
C. `(n(n+1))/(sqrt2)`
D. `(n(n-1))/(2)`

Answer

To find the sum of the first \( n \) terms of the series \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \ldots \), we first need to identify a pattern in the terms of the series.

### Step 1: Identify the Terms
The terms can be rewritten as follows:
- \( \sqrt{2} = \sqrt{2 \cdot 1} \)
- \( \sqrt{8} = \sqrt{2 \cdot 4} = \sqrt{2 \cdot 2^2} \)
- \( \sqrt{18} = \sqrt{2 \cdot 9} = \sqrt{2 \cdot 3^2} \)
- \( \sqrt{32} = \sqrt{2 \cdot 16} = \sqrt{2 \cdot 4^2} \)

This suggests a pattern where the \( n \)-th term can be expressed as:
\[ \sqrt{2n(n+1)} \]

### Step 2: General Term
The general term of the series can be expressed as:
\[ T_n = \sqrt{2n(n+1)} \]

### Step 3: Sum of the First \( n \) Terms
To find the sum of the first \( n \) terms, we sum the general term:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \sqrt{2k(k+1)} \]

### Step 4: Simplifying the Sum
We can factor out \( \sqrt{2} \):
\[ S_n = \sqrt{2} \sum_{k=1}^{n} \sqrt{k(k+1)} \]

### Step 5: Further Simplification
Using the identity \( \sqrt{k(k+1)} = \sqrt{k^2 + k} \), we can express the sum:
\[ S_n = \sqrt{2} \sum_{k=1}^{n} \sqrt{k^2 + k} \]

### Step 6: Final Expression
The exact evaluation of \( \sum_{k=1}^{n} \sqrt{k(k+1)} \) can be complex, but for practical purposes, we can use numerical methods or approximations to evaluate this sum for specific values of \( n \).

### Conclusion
Thus, the sum of the first \( n \) terms of the series \( \sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \ldots \) can be expressed as:
\[ S_n = \sqrt{2} \sum_{k=1}^{n} \sqrt{k(k+1)} \]