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Question Number 142647 by iloveisrael last updated on 03/Jun/21

$$\frac{1}{2018}-\frac{2}{2018}+\frac{3}{2018}-\frac{4}{2018}+...-\frac{2016}{2018}+\frac{2017}{2018}=?$$

Answered by MJS_new last updated on 03/Jun/21

$$\underset{j=1}{\sum ^{n+1}}(2j+1)-\underset{j=1}{\sum ^n}\left(2j\right)=n+1$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{1}{2}$$

Answered by Olaf_Thorendsen last updated on 03/Jun/21

$$\mathrm{S}=\underset{k=0}{\sum ^{1008}}(-\frac{2k}{2018}+\frac{2k+1}{2018})$$$$\mathrm{S}=\frac{1}{2018}\underset{k=0}{\sum ^{1008}}1=\frac{1}{2018}\times 1009=\frac{1}{2}$$

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