What is the value of $\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+...$ upto n terms?
Answer & explanation
Correct answer: option 4
$\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+...$ upto n terms
= ( 1 + 1 + 1 + ........upto n terms ) + (-\(\frac{1}{n}\) -\(\frac{2}{n}\) -\(\frac{3}{n}\) - ...... upto n terms )
= n - \(\frac{1}{n}\) ( 1 + 2 + 3 -......upto n terms )
= n - \(\frac{1}{n}\) ( \(\frac{n}{2}\) ( 1+ ( n - 1 ) 1 )
( Sn = \(\frac{n}{2}\) ( 2a + ( n - 1 ) d )
= n - \(\frac{1}{2}\) ( 2 + ( n - 1 ) 1 )
= n - \(\frac{1}{2}\) ( n + 1 )
= \(\frac{ n - 1}{2}\)
The correct answer is Option (4) → $\frac{n-1}{2}$