The sum of n terms of the following series `1+(1+x)+(1+x+x^2)+....` will be

Options

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

Answer

To find the sum of the first \( n \) terms of the series \( 1 + (1 + x) + (1 + x + x^2) + \ldots \), we can follow these steps:

### Step 1: Identify the \( n \)-th term of the series
The \( n \)-th term \( T_n \) of the series can be expressed as:
\[ T_n = 1 + x + x^2 + \ldots + x^{n-1} \]
This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = x \).

### Step 2: Use the formula for the sum of a geometric series
The sum \( S \) of the first \( n \) terms of a geometric series can be calculated using the formula:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
For our series, substituting \( a = 1 \) and \( r = x \) gives:
\[ T_n = \frac{1(1 - x^n)}{1 - x} = \frac{1 - x^n}{1 - x} \]

### Step 3: Find the total sum of the series up to \( n \) terms
Now, we need to find the sum of the first \( n \) terms of the series:
\[ S = T_1 + T_2 + T_3 + \ldots + T_n \]
Substituting the expression for \( T_k \) (where \( k \) varies from 1 to \( n \)):
\[ S = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{1 - x^k}{1 - x} \]
This can be simplified as:
\[ S = \frac{1}{1 - x} \sum_{k=1}^{n} (1 - x^k) \]

### Step 4: Simplify the summation
The summation can be separated:
\[ S = \frac{1}{1 - x} \left( \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} x^k \right) \]
The first summation \( \sum_{k=1}^{n} 1 = n \) and the second summation \( \sum_{k=1}^{n} x^k \) is again a geometric series:
\[ \sum_{k=1}^{n} x^k = \frac{x(1 - x^n)}{1 - x} \]
Thus, we have:
\[ S = \frac{1}{1 - x} \left( n - \frac{x(1 - x^n)}{1 - x} \right) \]

### Step 5: Combine and simplify
Now, substituting back, we get:
\[ S = \frac{n(1 - x) - x(1 - x^n)}{(1 - x)^2} \]
This simplifies to:
\[ S = \frac{n - nx - x + x^{n+1}}{(1 - x)^2} \]

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