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Question

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

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Solution

## On simplifying the given series, we get: $\left(1-\frac{1}{n}\right)+\left(1-\frac{2}{n}\right)+\left(1-\frac{3}{n}\right)+...n\mathrm{terms}\phantom{\rule{0ex}{0ex}}=\left(1+1+1+...n\mathrm{terms}\right)-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right)\phantom{\rule{0ex}{0ex}}=n-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Here},\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+...+\frac{n}{n}\right)\mathrm{is}\mathrm{an}\mathrm{AP}\mathrm{whose}\mathrm{first}\mathrm{term}\mathrm{is}\frac{1}{\mathrm{n}}\mathrm{and}\mathrm{the}\mathrm{common}\mathrm{difference}\mathrm{is}\left(\frac{2}{n}-\frac{1}{n}\right)=\frac{1}{n}.\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{sum}\mathrm{of}n\mathrm{terms}\mathrm{of}\mathrm{an}\mathrm{AP}\mathrm{is}\mathrm{given}\mathrm{by}\phantom{\rule{0ex}{0ex}}Sn=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}=n-\left[\frac{n}{2}\left\{2×\left(\frac{1}{n}\right)+\left(n-1\right)×\left(\frac{1}{n}\right)\right\}\right]\phantom{\rule{0ex}{0ex}}=n-\left[\frac{n}{2}\left[\left(\frac{2}{n}\right)+\left(\frac{n-1}{n}\right)\right]\right]=n-\left\{\frac{n}{2}\left(\frac{n+1}{n}\right)\right\}\phantom{\rule{0ex}{0ex}}=n-\left(\frac{n+1}{2}\right)=\frac{n-1}{2}\phantom{\rule{0ex}{0ex}}$

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