Sum of n terms of the series `(1)/(3) + (5)/(9) + (19)/(27) + (65)/(81)+…` is

Options

A. `2^(n) - 1`
B. `3^(n) - 1`
C. `n - ((3^(n) - 2^(n)))/(2^(n))`
D. `n - (2 (3^(n) - 2^(n)))/(3^(n))`

Answer

To find the sum of the first \( n \) terms of the series

\[ \frac{1}{3} + \frac{5}{9} + \frac{19}{27} + \frac{65}{81} + \ldots \]

we can analyze the series and derive a formula for the sum.

### Step 1: Identify the pattern in the numerators and denominators

The denominators of the series are powers of 3:
- \( 3^1, 3^2, 3^3, 3^4, \ldots \)

The numerators appear to follow a pattern:
- The numerators are \( 1, 5, 19, 65, \ldots \)

### Step 2: Find a formula for the numerators

We can observe that:
- \( 1 = 2^0 + 1 \)
- \( 5 = 2^2 + 1 \)
- \( 19 = 2^4 + 3 \)
- \( 65 = 2^6 + 1 \)

This suggests that the numerators can be expressed in terms of powers of 2. We can derive a general formula for the \( n \)-th term of the numerator.

### Step 3: Express the \( n \)-th term of the series

The \( n \)-th term of the series can be expressed as:

\[ \frac{2^{n+1} - 1}{3^n} \]

### Step 4: Write the sum of the first \( n \) terms

The sum \( S_n \) of the first \( n \) terms can be expressed as:

\[ S_n = \sum_{k=1}^{n} \frac{2^{k+1} - 1}{3^k} \]

This can be split into two separate sums:

\[ S_n = \sum_{k=1}^{n} \frac{2^{k+1}}{3^k} - \sum_{k=1}^{n} \frac{1}{3^k} \]

### Step 5: Calculate each sum

1. **First sum**:

\[ \sum_{k=1}^{n} \frac{2^{k+1}}{3^k} = 2 \sum_{k=1}^{n} \left(\frac{2}{3}\right)^k = 2 \cdot \frac{\frac{2}{3}(1 - (\frac{2}{3})^n)}{1 - \frac{2}{3}} = 2 \cdot \frac{2}{3} \cdot \frac{1 - (\frac{2}{3})^n}{\frac{1}{3}} = 4(1 - (\frac{2}{3})^n) \]

2. **Second sum**:

\[ \sum_{k=1}^{n} \frac{1}{3^k} = \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} = \frac{1}{3} \cdot \frac{1 - (\frac{1}{3})^n}{\frac{2}{3}} = \frac{1}{2}(1 - (\frac{1}{3})^n) \]

### Step 6: Combine the results

Now substituting back into the expression for \( S_n \):

\[ S_n = 4(1 - (\frac{2}{3})^n) - \frac{1}{2}(1 - (\frac{1}{3})^n) \]

### Step 7: Simplify

Combine the terms:

\[ S_n = 4 - 4(\frac{2}{3})^n - \frac{1}{2} + \frac{1}{2}(\frac{1}{3})^n \]

\[ S_n = \frac{8}{2} - \frac{1}{2} - 4(\frac{2}{3})^n + \frac{1}{2}(\frac{1}{3})^n \]

\[ S_n = \frac{7}{2} - 4(\frac{2}{3})^n + \frac{1}{2}(\frac{1}{3})^n \]

### Final Result

Thus, the sum of the first \( n \) terms of the series is:

\[ S_n = \frac{7}{2} - 4 \left(\frac{2}{3}\right)^n + \frac{1}{2} \left(\frac{1}{3}\right)^n \]