`sum_(k =1)^(n) k(1 + 1/n)^(k -1)` =

Options

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

Answer

To find the sum \( S = \sum_{k=1}^{n} k \left(1 + \frac{1}{n}\right)^{k-1} \), we can use a systematic approach.

### Step-by-Step Solution:

1. **Define the Sum**:
\[ S = \sum_{k=1}^{n} k \left(1 + \frac{1}{n}\right)^{k-1} \]

2. **Identify the Common Ratio**:
Let \( t = 1 + \frac{1}{n} \). Thus, we can rewrite the sum as:
\[ S = \sum_{k=1}^{n} k t^{k-1} \]

3. **Use the Formula for the Sum of a Geometric Series**:
The sum \( \sum_{k=1}^{n} k x^{k-1} \) can be derived using the formula:
\[ \sum_{k=1}^{n} k x^{k-1} = x \frac{d}{dx} \left( \sum_{k=0}^{n} x^k \right) \]
where \( \sum_{k=0}^{n} x^k = \frac{1 - x^{n+1}}{1 - x} \).

4. **Differentiate the Geometric Series**:
Differentiate \( \sum_{k=0}^{n} x^k \):
\[ \frac{d}{dx} \left( \frac{1 - x^{n+1}}{1 - x} \right) = \frac{(n+1)x^n(1-x) + (1-x^{n+1})}{(1-x)^2} \]

5. **Substituting Back**:
Now substituting \( x = t \):
\[ \sum_{k=1}^{n} k t^{k-1} = t \cdot \frac{(n+1)t^n(1-t) + (1-t^{n+1})}{(1-t)^2} \]

6. **Final Expression**:
This gives:
\[ S = \frac{t(1 - t^{n+1}) + (n+1)t^{n+1}(1 - t)}{(1 - t)^2} \]

7. **Substituting \( t = 1 + \frac{1}{n} \)**:
Substitute back \( t = 1 + \frac{1}{n} \) into the expression to get the final result.

### Final Result:
After simplifying, we find:
\[ S = \frac{(n+1)^{n+1} - n^n}{n} \]