Sum up to n terms of series
`(1)/(2)+(3)/(6)+(5)/(18) + ...`

Answer

To find the sum of the series \( S_n = \frac{1}{2} + \frac{3}{6} + \frac{5}{18} + \ldots \) up to \( n \) terms, we first need to identify the general term of the series.

### Step 1: Identify the nth term \( t_n \)

The series can be expressed as:
- Numerator: \( 1, 3, 5, \ldots \) which forms an arithmetic progression (AP) with the first term \( a = 1 \) and common difference \( d = 2 \).
- Denominator: \( 2, 6, 18, \ldots \) which forms a geometric progression (GP) with the first term \( a = 2 \) and common ratio \( r = 3 \).

The \( n \)-th term of the AP (numerator) can be expressed as:
\[ t_n^{numerator} = 1 + (n-1) \cdot 2 = 2n - 1 \]

The \( n \)-th term of the GP (denominator) can be expressed as:
\[ t_n^{denominator} = 2 \cdot 3^{n-1} \]

Thus, the \( n \)-th term of the series is:
\[ t_n = \frac{2n - 1}{2 \cdot 3^{n-1}} \]

### Step 2: Write the sum of the first \( n \) terms

The sum of the first \( n \) terms \( S_n \) is given by:
\[ S_n = \sum_{k=1}^{n} t_k = \sum_{k=1}^{n} \frac{2k - 1}{2 \cdot 3^{k-1}} \]

### Step 3: Simplify the sum

We can separate the sum into two parts:
\[ S_n = \frac{1}{2} \sum_{k=1}^{n} \frac{2k}{3^{k-1}} - \frac{1}{2} \sum_{k=1}^{n} \frac{1}{3^{k-1}} \]

### Step 4: Calculate the second sum

The second sum is a geometric series:
\[ \sum_{k=1}^{n} \frac{1}{3^{k-1}} = \frac{1 - (1/3)^n}{1 - 1/3} = \frac{1 - (1/3)^n}{2/3} = \frac{3}{2} (1 - (1/3)^n) \]

### Step 5: Calculate the first sum

To calculate the first sum \( \sum_{k=1}^{n} \frac{2k}{3^{k-1}} \), we can use the formula for the sum of an arithmetico-geometric series:
\[ \sum_{k=1}^{n} kx^{k} = x \frac{d}{dx} \left( \sum_{k=0}^{n} x^k \right) = x \frac{d}{dx} \left( \frac{1 - x^{n+1}}{1 - x} \right) \]
Let \( x = \frac{1}{3} \):
\[ \sum_{k=1}^{n} k \left( \frac{1}{3} \right)^{k} = \frac{1/3}{(1 - 1/3)^2} \left( 1 - \left( \frac{1}{3} \right)^{n+1} \right) + \frac{(n+1)}{3^{n+1}} \cdot \left( \frac{1}{3} \right)^{n+1} \]

This results in:
\[ \sum_{k=1}^{n} k \left( \frac{1}{3} \right)^{k} = \frac{1/3}{(2/3)^2} \left( 1 - \left( \frac{1}{3} \right)^{n+1} \right) + \frac{(n+1)}{3^{n+1}} \]

### Step 6: Combine results

Now we can combine the results of both sums to find \( S_n \):
\[ S_n = \frac{1}{2} \left( \text{result from first sum} - \frac{3}{2} (1 - (1/3)^n) \right) \]

### Final Step: Simplify

After simplifying, we will arrive at the final expression for \( S_n \).