Find the sum to n terms :
`1+ 2x + 3x^2 + 4x^3 + … ,xne 1` :

Options

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

Answer

To find the sum to \( n \) terms of the series \( 1 + 2x + 3x^2 + 4x^3 + \ldots \), we can follow these steps:

### Step 1: Define the Series
Let \( S_n = 1 + 2x + 3x^2 + 4x^3 + \ldots + nx^{n-1} \).

### Step 2: Multiply the Series by \( x \)
Now, multiply the entire series \( S_n \) by \( x \):
\[ x S_n = x + 2x^2 + 3x^3 + 4x^4 + \ldots + nx^n \]

### Step 3: Subtract the Two Equations
Now, subtract the second equation from the first:
\[ S_n - x S_n = (1 + 2x + 3x^2 + 4x^3 + \ldots + nx^{n-1}) - (x + 2x^2 + 3x^3 + 4x^4 + \ldots + nx^n) \]
This simplifies to:
\[ S_n - x S_n = 1 + (2x - x) + (3x^2 - 2x^2) + (4x^3 - 3x^3) + \ldots + (nx^{n-1} - nx^n) \]
\[ S_n - x S_n = 1 + x + x^2 + x^3 + \ldots + x^{n-1} \]

### Step 4: Recognize the Geometric Series
The right-hand side is a geometric series with first term \( 1 \) and common ratio \( x \):
\[ 1 + x + x^2 + x^3 + \ldots + x^{n-1} = \frac{1 - x^n}{1 - x} \quad \text{(for } x \neq 1\text{)} \]

### Step 5: Substitute Back
Now, we can substitute this back into our equation:
\[ S_n(1 - x) = \frac{1 - x^n}{1 - x} \]

### Step 6: Solve for \( S_n \)
Now, solve for \( S_n \):
\[ S_n = \frac{1 - x^n}{(1 - x)(1 - x)} = \frac{1 - x^n}{(1 - x)^2} \]

### Final Result
Thus, the sum to \( n \) terms of the series \( 1 + 2x + 3x^2 + 4x^3 + \ldots \) is:
\[ S_n = \frac{1 - x^{n}}{(1 - x)^2} \]