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

 (1/(2018))−(2/(2018))+(3/(2018))−(4/(2018))+...−((2016)/(2018))+((2017)/(2018))=?

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

Answered by MJS_new last updated on 03/Jun/21

Σ_(j=1) ^(n+1) (2j+1)−Σ_(j=1) ^n (2j)=n+1  ⇒ answer is (1/2)

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

Answered by Olaf_Thorendsen last updated on 03/Jun/21

S = Σ_(k=0) ^(1008) (−((2k)/(2018))+((2k+1)/(2018)))  S = (1/(2018))Σ_(k=0) ^(1008) 1 = (1/(2018))×1009 = (1/2)

$$\mathrm{S}\:=\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{1008}} {\sum}}\left(−\frac{\mathrm{2}{k}}{\mathrm{2018}}+\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{2018}}\right) \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{2018}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{1008}} {\sum}}\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{2018}}×\mathrm{1009}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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