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 separate parts: one part containing the terms with \( x \) and the other part containing the constant terms.

### Step 1: Identify the two parts of the series
The series can be rewritten as:
1. Terms with \( x \): \( x, 3x, 5x, \ldots \)
2. Constant terms: \( \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, \ldots \)

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

The \( n \)-th term of an AP is given by:
\[ a_n = a + (n - 1)d \]
For our series, the \( n \)-th term is:
\[ a_n = x + (n - 1)(2x) = x + 2(n - 1)x = (2n - 1)x \]

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

### Step 4: Analyze the second part (constant terms)
The second part forms a geometric progression (GP):
- First term \( a = \frac{1}{2} \)
- Common ratio \( r = \frac{-\frac{1}{6}}{\frac{1}{2}} = -\frac{1}{3} \)

### Step 5: Calculate the sum of the second part
The sum \( S_n \) of the first \( n \) terms of a GP is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
Substituting \( n = 20 \), \( a = \frac{1}{2} \), and \( r = -\frac{1}{3} \):
\[ S_{20} = \frac{1}{2} \frac{1 - (-\frac{1}{3})^{20}}{1 - (-\frac{1}{3})} = \frac{1}{2} \frac{1 - \frac{1}{3^{20}}}{1 + \frac{1}{3}} = \frac{1}{2} \frac{1 - \frac{1}{3^{20}}}{\frac{4}{3}} = \frac{3}{8} \left(1 - \frac{1}{3^{20}}\right) \]

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

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