Sum of series `n+(n-1)x+(n-2)x^2+........+2x^(n-2)+x^(n-1).`

Answer

To find the sum of the series \( S_n = n + (n-1)x + (n-2)x^2 + \ldots + 2x^{n-2} + x^{n-1} \), we can follow these steps:

### Step 1: Define the series
Let:
\[ S_n = n + (n-1)x + (n-2)x^2 + \ldots + 2x^{n-2} + x^{n-1} \]

### Step 2: Multiply the series by \( x \)
Now, multiply the entire series \( S_n \) by \( x \):
\[ x S_n = n x + (n-1)x^2 + (n-2)x^3 + \ldots + 2x^{n-1} + x^n \]

### Step 3: Subtract the second equation from the first
Now, we subtract the equation \( x S_n \) from \( S_n \):
\[ S_n - x S_n = n + (n-1)x + (n-2)x^2 + \ldots + 2x^{n-2} + x^{n-1} - (n x + (n-1)x^2 + (n-2)x^3 + \ldots + 2x^{n-1} + x^n) \]

### Step 4: Simplify the left-hand side
This simplifies to:
\[ (1 - x) S_n = n - (x + x^2 + x^3 + \ldots + x^n) \]

### Step 5: Recognize the geometric series
The series \( x + x^2 + x^3 + \ldots + x^n \) is a geometric series with first term \( x \) and common ratio \( x \). The sum of this series can be calculated using the formula for the sum of a geometric series:
\[ \text{Sum} = \frac{a(r^n - 1)}{r - 1} \]
where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = x \) and \( r = x \):
\[ x + x^2 + x^3 + \ldots + x^n = x \frac{x^n - 1}{x - 1} \]

### Step 6: Substitute back into the equation
Substituting this back into our equation gives:
\[ (1 - x) S_n = n - x \frac{x^n - 1}{x - 1} \]

### Step 7: Solve for \( S_n \)
Now, we can solve for \( S_n \):
\[ S_n = \frac{n - x \frac{x^n - 1}{x - 1}}{1 - x} \]

### Step 8: Simplify the expression
To simplify further:
\[ S_n = \frac{n(1-x) - x(x^n - 1)}{(1-x)(x-1)} \]
This can be rearranged to:
\[ S_n = \frac{n(1-x) + x(x^n - 1)}{(1-x)^2} \]

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