Find the sum of n terms of the series `(x+y)+(x^(2)+2y)+(x^(3)+3y)+...`

Explanation

To find the sum of the first \( n \) terms of the series \( (x+y) + (x^2 + 2y) + (x^3 + 3y) + \ldots \), we can break the series into two separate parts: one part involving the powers of \( x \) and the other part involving the multiples of \( y \).

### Step-by-Step Solution:

1. **Identify the Series Components**:
The series can be rewritten as:
\[ (x + x^2 + x^3 + \ldots + x^n) + (y + 2y + 3y + \ldots + ny) \]
This separates the series into two distinct series.

2. **Sum of the First Series**:
The first series is a geometric series:
\[ S_1 = x + x^2 + x^3 + \ldots + x^n \]
This is a geometric progression (GP) where:
- First term \( A = x \)
- Common ratio \( R = x \)
- Number of terms \( n \)

The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = A \frac{R^n - 1}{R - 1} \]
Substituting the values:
\[ S_1 = x \frac{x^n - 1}{x - 1} \]

3. **Sum of the Second Series**:
The second series is an arithmetic series:
\[ S_2 = y + 2y + 3y + \ldots + ny \]
This is an arithmetic progression (AP) where:
- First term \( A = y \)
- Common difference \( D = y \)
- Number of terms \( n \)

The formula for the sum of the first \( n \) terms of an arithmetic series is:
\[ S_n = \frac{n}{2} \left(2A + (n-1)D\right) \]
Substituting the values:
\[ S_2 = \frac{n}{2} \left(2y + (n-1)y\right) = \frac{n}{2} \left(2y + ny - y\right) = \frac{n}{2} \left(n + 1\right)y \]

4. **Combine the Two Sums**:
Now, we combine the sums from both series to find the total sum \( S \):
\[ S = S_1 + S_2 = x \frac{x^n - 1}{x - 1} + \frac{n(n + 1)}{2} y \]

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