$$\lparen1-\frac{1}{n}\rparen+\lparen1-\frac{2}{n}\rparen+\lparen1-\frac{3}{n}\rparen+ ..$$​ up to n terms will result as:

Correct option is D

Given:

The series is:

​$$\left( 1 - \frac{1}{n} \right) + \left( 1 - \frac{2}{n} \right) + \left( 1 - \frac{3}{n} \right) + \dots \text{ up to } n \text{ terms.}$$​​

We need to find the sum of the series.

Formula Used:

For the given series, the sum is:

​$$S_n = \left( \sum_{k=1}^{n} 1 \right) - \left( \sum_{k=1}^{n} \frac{k}{n} \right)$$​​

Solution:

​$$\sum_{k=1}^{n} 1 = n$$​​

​$$\sum_{k=1}^{n} \frac{k}{n} = \frac{1}{n} \sum_{k=1}^{n} k = \frac{1}{n} \times \frac{n(n+1)}{2} = \frac{n+1}{2}$$​​

​$$S_n = n - \frac{n+1}{2}$$​​

​$$S_n = \frac{2n - (n+1)}{2} = \frac{n-1}{2}$$

Option (D) is right.