If `S = (2)/(3) + (8)/(9) + (26)/(27) + (30)/(81)+….+n` terms, then the value of S is equal to

Options

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

Answer

To find the value of \( S = \frac{2}{3} + \frac{8}{9} + \frac{26}{27} + \frac{30}{81} + \ldots \) for \( n \) terms, we can analyze the pattern of the terms.

### Step 1: Identify the pattern in the terms
The terms can be rewritten as:
- First term: \( \frac{2}{3} = 1 - \frac{1}{3} \)
- Second term: \( \frac{8}{9} = 1 - \frac{1}{9} \)
- Third term: \( \frac{26}{27} = 1 - \frac{1}{27} \)
- Fourth term: \( \frac{30}{81} = 1 - \frac{1}{81} \)

From this, we can see that the \( n \)-th term can be expressed as:
\[ \text{n-th term} = 1 - \frac{1}{3^n} \]

### Step 2: Write the sum \( S \)
The sum \( S \) can be expressed as:
\[ S = \sum_{n=1}^{N} \left( 1 - \frac{1}{3^n} \right) \]
This can be split into two separate sums:
\[ S = \sum_{n=1}^{N} 1 - \sum_{n=1}^{N} \frac{1}{3^n} \]

### Step 3: Calculate the first sum
The first sum, \( \sum_{n=1}^{N} 1 \), simply counts the number of terms:
\[ \sum_{n=1}^{N} 1 = N \]

### Step 4: Calculate the second sum
The second sum \( \sum_{n=1}^{N} \frac{1}{3^n} \) is a geometric series with first term \( a = \frac{1}{3} \) and common ratio \( r = \frac{1}{3} \). The formula for the sum of the first \( N \) terms of a geometric series is:
\[ S_n = a \frac{1 - r^N}{1 - r} \]
Substituting the values:
\[ \sum_{n=1}^{N} \frac{1}{3^n} = \frac{\frac{1}{3} \left( 1 - \left( \frac{1}{3} \right)^N \right)}{1 - \frac{1}{3}} = \frac{\frac{1}{3} \left( 1 - \frac{1}{3^N} \right)}{\frac{2}{3}} = \frac{1}{2} \left( 1 - \frac{1}{3^N} \right) \]

### Step 5: Combine the results
Now we can substitute back into the expression for \( S \):
\[ S = N - \frac{1}{2} \left( 1 - \frac{1}{3^N} \right) \]
This simplifies to:
\[ S = N - \frac{1}{2} + \frac{1}{6^N} \]

### Final Expression
Thus, the value of \( S \) is:
\[ S = N - \frac{1}{2} \left( 1 - \frac{1}{3^N} \right) \]