The sum of the series `1+4/3 +10/9+28/27+.....` upto n terms is

Options

A. `n-(1)/(3)+(1)/(3.2^(n-1))`
B. `(7)/(6)n+(1)/(6)+(1)/(3.2^(n-1))`
C. `(5)/(3)n-(7)/(6)+(1)/(2.3^(n-1))`
D. `n+(1)/(2)-(1)/(2.3^(n-1))`

Explanation

To find the sum of the series \( S = 1 + \frac{4}{3} + \frac{10}{9} + \frac{28}{27} + \ldots \) up to \( n \) terms, we can break down the series into a more manageable form.

### Step 1: Rewrite the terms of the series
The terms can be rewritten as:
\[ S = 1 + \left(1 + \frac{1}{3}\right) + \left(1 + \frac{1}{9}\right) + \left(1 + \frac{1}{27}\right) + \ldots \]
This can be expressed as:
\[ S = n + \left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\right) \]
Here, \( n \) is the number of terms, and the remaining part is a geometric series.

### Step 2: Identify the geometric series
The series \( \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \) is a geometric series where:
- First term \( a = \frac{1}{3} \)
- Common ratio \( r = \frac{1}{3} \)

### Step 3: Sum of the geometric series
The sum \( S_G \) of a geometric series can be calculated using the formula:
\[ S_G = \frac{a(1 - r^n)}{1 - r} \]
Substituting the values we have:
\[ S_G = \frac{\frac{1}{3}(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} = \frac{\frac{1}{3}(1 - \frac{1}{3^n})}{\frac{2}{3}} = \frac{1 - \frac{1}{3^n}}{2} \]

### Step 4: Combine the sums
Now, we combine the sum of the geometric series with the \( n \) terms:
\[ S = n + S_G = n + \frac{1 - \frac{1}{3^n}}{2} \]
This simplifies to:
\[ S = n + \frac{1}{2} - \frac{1}{2 \cdot 3^n} \]

### Final Result
Thus, the sum of the series up to \( n \) terms is:
\[ S = n + \frac{1}{2} - \frac{1}{2 \cdot 3^n} \]