$\left(1-\dfrac{1}{n}\right)$+$\left(1-\dfrac{2}{n}\right)$+$\left(1-\dfrac{3}{n}\right)$+.....up to n terms=?
$ \dfrac{1}{2} $n
$ \dfrac{1}{2} $(n - 1)
$ \dfrac{1}{2} $n(n - 1)
None of these
Explanation:
Given sum
= [1 + 1 + 1 + ... to n terms]-$(\dfrac{1}{n}+\dfrac{2}{n}+\dfrac{3}{n}+ ... $)to n terms
= $ n $ -$\dfrac{ n }{2}$$\left(\dfrac{1}{ n }\right)$+ 1 [ Ref: $ n $th terms = $\left( n / n\right )$ = 1]
= $ n $ -$\dfrac{ n + 1}{2} $
=$ \dfrac{1}{2} \left( n - 1\right)$