Question

Find the sum of the following:
$\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+...\mathrm{upto}n\mathrm{terms}$.

Answer 1

On simplifying the given series, we get:

$$\begin{gathered} \left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+...n\mathrm{terms} \\ =(1+1+1+...n\mathrm{terms})-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right) \\ =n-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right) \\ \mathrm{Here},\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right)\mathrm{is}\mathrm{an}\mathrm{AP}\mathrm{whose}\mathrm{first}\mathrm{term}\mathrm{is}\frac{1}{\mathrm{n}}\mathrm{and}\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{is}\left(\frac{2}{n}-\frac{1}{n}\right)=\frac{1}{n}. \\ \mathrm{The}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}\mathrm{is}\mathrm{given}\mathrm{by} \\ Sn=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ =n-\left[\frac{n}{2}\left\{2\times \left({\displaystyle \frac{1}{n}}\right)+(n-1)\times \left(\frac{1}{n}\right)\right\}\right] \\ =n-\left[\frac{n}{2}\left[\left(\frac{2}{n}\right)+\left(\frac{n-1}{n}\right)\right]\right]=n-\left\{\frac{n}{2}\left(\frac{n+1}{n}\right)\right\} \\ =n-\left(\frac{n+1}{2}\right)=\frac{n-1}{2} \end{gathered}$$