If the `r^(th)` term of a series is `1 + x + x^2 + .......+ x^(r-1)` , then the sum of the first n terms is

Options

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

Answer

To find the sum of the first \( n \) terms of the series where the \( r^{th} \) term is given by

\[ a_r = 1 + x + x^2 + \ldots + x^{r-1}, \]

we can follow these steps:

### Step 1: Identify the \( r^{th} \) term as a geometric series
The \( r^{th} \) term can be recognized as a geometric series with:
- First term \( a = 1 \)
- Common ratio \( r = x \)
- Number of terms \( r \)

The formula for the sum of the first \( r \) terms of a geometric series is given by:

\[ S_r = \frac{a(1 - r^n)}{1 - r} \]

Substituting the values, we have:

\[ a_r = \frac{1(1 - x^r)}{1 - x} = \frac{1 - x^r}{1 - x}. \]

### Step 2: Find the sum of the first \( n \) terms
The sum of the first \( n \) terms, denoted as \( S_n \), can be expressed as:

\[ S_n = \sum_{r=1}^{n} a_r = \sum_{r=1}^{n} \frac{1 - x^r}{1 - x}. \]

### Step 3: Separate the summation
We can separate the summation into two parts:

\[ S_n = \frac{1}{1 - x} \sum_{r=1}^{n} (1 - x^r) = \frac{1}{1 - x} \left( \sum_{r=1}^{n} 1 - \sum_{r=1}^{n} x^r \right). \]

### Step 4: Calculate the summations
1. The first summation \( \sum_{r=1}^{n} 1 = n \).
2. The second summation \( \sum_{r=1}^{n} x^r \) is also a geometric series, which can be calculated as:

\[ \sum_{r=1}^{n} x^r = x + x^2 + \ldots + x^n = \frac{x(1 - x^n)}{1 - x}. \]

### Step 5: Substitute back into the equation
Now substituting these results back into the equation for \( S_n \):

\[ S_n = \frac{1}{1 - x} \left( n - \frac{x(1 - x^n)}{1 - x} \right). \]

### Step 6: Simplify the expression
This can be simplified further:

\[ S_n = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2}. \]

Expanding this gives:

\[ S_n = \frac{n - nx - x + x^{n+1}}{(1 - x)^2}. \]

### Final Result
Thus, the sum of the first \( n \) terms is:

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