Sum to n terms of the series `:1 + 2 (1 + (1)/(n)) + 3 (1 + (1)/(n )) ^(2) + ………`

Answer

To find the sum to n terms of the series \( S_n = 1 + 2 \left(1 + \frac{1}{n}\right) + 3 \left(1 + \frac{1}{n}\right)^2 + \ldots \), we can follow these steps:

### Step 1: Define the series
Let’s denote the series as:
\[ S_n = 1 + 2 \left(1 + \frac{1}{n}\right) + 3 \left(1 + \frac{1}{n}\right)^2 + \ldots + n \left(1 + \frac{1}{n}\right)^{n-1} \]

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

### Step 3: Set up the equation
Now, we can set up the equation:
\[ S_n \left(1 + \frac{1}{n}\right) - S_n = 1 + \left(1 + \frac{1}{n}\right) + \left(1 + \frac{1}{n}\right)^2 + \ldots + \left(1 + \frac{1}{n}\right)^{n-1} \]

### Step 4: Simplify the left-hand side
The left-hand side simplifies to:
\[ S_n \cdot \frac{1}{n} = S_n \left(1 + \frac{1}{n}\right) - S_n = S_n \cdot \frac{1}{n} \]

### Step 5: Recognize the right-hand side as a geometric series
The right-hand side is a geometric series with first term \( 1 \) and common ratio \( \left(1 + \frac{1}{n}\right) \):
\[ \text{Sum} = \frac{1 - \left(1 + \frac{1}{n}\right)^n}{1 - \left(1 + \frac{1}{n}\right)} \]

### Step 6: Calculate the sum of the geometric series
The sum of the geometric series can be simplified:
\[ \text{Sum} = \frac{1 - \left(1 + \frac{1}{n}\right)^n}{-\frac{1}{n}} = -n \left(1 - \left(1 + \frac{1}{n}\right)^n\right) \]

### Step 7: Substitute back into the equation
Substituting this back into our equation gives:
\[ S_n \cdot \frac{1}{n} = -n \left(1 - \left(1 + \frac{1}{n}\right)^n\right) \]

### Step 8: Solve for \( S_n \)
Multiplying both sides by \( n \):
\[ S_n = -n^2 \left(1 - \left(1 + \frac{1}{n}\right)^n\right) \]

### Step 9: Simplify further
As \( n \) approaches infinity, \( \left(1 + \frac{1}{n}\right)^n \) approaches \( e \). However, for our calculation, we keep it as is:
\[ S_n = n^2 \left(1 - \left(1 + \frac{1}{n}\right)^n\right) \]

### Final Result
Thus, the sum to n terms of the series is:
\[ S_n = n^2 \]