`(1-1/n) + (1 - 2/n) + (1 - 3/n)+ ……` upto n terms = ?

Options

A. `1/2n`
B. `1/2(n-1)`
C. `1/2 n (n-1)`
D. none of these

Answer

To find the sum of the series \( (1 - \frac{1}{n}) + (1 - \frac{2}{n}) + (1 - \frac{3}{n}) + \ldots \) up to \( n \) terms, we can follow these steps:

### Step 1: Rewrite the Series
The series can be rewritten as:
\[ S_n = \sum_{k=1}^{n} \left(1 - \frac{k}{n}\right) \]
This means we need to sum up \( n \) terms of the form \( 1 - \frac{k}{n} \).

### Step 2: Split the Summation
We can split the summation into two parts:
\[ S_n = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{k}{n} \]
The first part, \( \sum_{k=1}^{n} 1 \), is simply \( n \) since we are adding \( 1 \) for each of the \( n \) terms.

### Step 3: Calculate the Second Summation
The second part can be simplified as:
\[ \sum_{k=1}^{n} \frac{k}{n} = \frac{1}{n} \sum_{k=1}^{n} k \]
We know that the sum of the first \( n \) natural numbers is given by the formula:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
Thus, we can substitute this into our equation:
\[ \sum_{k=1}^{n} \frac{k}{n} = \frac{1}{n} \cdot \frac{n(n + 1)}{2} = \frac{n + 1}{2} \]

### Step 4: Substitute Back into the Summation
Now substituting back into our expression for \( S_n \):
\[ S_n = n - \frac{n + 1}{2} \]

### Step 5: Simplify the Expression
Now we simplify:
\[ S_n = n - \frac{n + 1}{2} = \frac{2n}{2} - \frac{n + 1}{2} = \frac{2n - n - 1}{2} = \frac{n - 1}{2} \]

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

---