Find the sum of 20 terms of the series `(x+(1)/(2))+(3x-(1)/(6))+(5x+(1)/(18))+....`

Explanation

To find the sum of the first 20 terms of the series \( (x + \frac{1}{2}) + (3x - \frac{1}{6}) + (5x + \frac{1}{18}) + \ldots \), we will break the series into two parts: one that involves \( x \) and another that involves the constant terms.

### Step 1: Identify the two parts of the series
The series can be separated into two distinct parts:
1. The terms involving \( x \): \( x, 3x, 5x, \ldots \)
2. The constant terms: \( \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, \ldots \)

### Step 2: Analyze the first part (terms involving \( x \))
The first part forms an arithmetic progression (AP) with:
- First term \( a = x \)
- Common difference \( d = 3x - x = 2x \)

The \( n \)-th term of this AP can be expressed as:
\[ T_n = a + (n - 1)d = x + (n - 1)(2x) = x + 2(n - 1)x = (2n - 1)x \]

### Step 3: Calculate the sum of the first 20 terms of the AP
The sum of the first \( n \) terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
Substituting \( n = 20 \), \( a = x \), and \( d = 2x \):
\[ S_{20} = \frac{20}{2} \times \left(2x + (20 - 1)(2x)\right) \]
\[ = 10 \times \left(2x + 19 \cdot 2x\right) \]
\[ = 10 \times (2x + 38x) = 10 \times 40x = 400x \]

### Step 4: Analyze the second part (constant terms)
The constant terms form a series:
\[ \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, \ldots \]

To find the pattern, we can express the constant terms:
- The first term \( a = \frac{1}{2} \)
- The second term \( -\frac{1}{6} = -\frac{1}{2 \cdot 3} \)
- The third term \( \frac{1}{18} = \frac{1}{2 \cdot 3^2} \)

The general term can be expressed as:
\[ T_n = \frac{(-1)^{n-1}}{2 \cdot 3^{n-1}} \]

### Step 5: Calculate the sum of the first 20 terms of the constant series
This series is a geometric progression (GP) with:
- First term \( a = \frac{1}{2} \)
- Common ratio \( r = -\frac{1}{3} \)

The sum of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
Substituting \( n = 20 \), \( a = \frac{1}{2} \), and \( r = -\frac{1}{3} \):
\[ S_{20} = \frac{\frac{1}{2}(1 - (-\frac{1}{3})^{20})}{1 - (-\frac{1}{3})} \]
\[ = \frac{\frac{1}{2}(1 - \frac{1}{3^{20}})}{1 + \frac{1}{3}} = \frac{\frac{1}{2}(1 - \frac{1}{3^{20}})}{\frac{4}{3}} = \frac{3}{8}(1 - \frac{1}{3^{20}}) \]

### Step 6: Combine the sums of both parts
The total sum of the first 20 terms of the series is:
\[ S_{20} = S_{AP} + S_{GP} = 400x + \frac{3}{8}(1 - \frac{1}{3^{20}}) \]

### Final Answer
Thus, the sum of the first 20 terms of the series is:
\[ S_{20} = 400x + \frac{3}{8}(1 - \frac{1}{3^{20}}) \]