Question

The sum of ${}^{\prime }{n}^{\prime }$ terms of series $(1-\frac{1}{n})+(1-\frac{2}{n})+(1-\frac{3}{n})+.............$ will be:

Answer 1

Let,$S_n=(1-\frac{1}{n})+(1-\frac{2}{n})+(1-\frac{3}{n})+.............n\text{terms}$ .

$\therefore S_n=n-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+....+\frac{n}{n}\right)$

$=n-\left(\frac{1+2+3+.......+n}{n}\right)$

$=n-\frac{n(n+1)}{2n}$ [Since, $1+2+3+......+n=\frac{n(n+1)}{2}$]

$=n-\frac{n+1}{2}$

$=\frac{n-1}{2}$