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Question Number 126963 by ZiYangLee last updated on 25/Dec/20

$$\mathrm{If}\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+\frac{1}{7\times 9}+\dots +\frac{1}{2017\times 2019}=\frac{x}{y}$$$$\mathrm{where}\frac{x}{y}\mathrm{is}\mathrm{in}\mathrm{its}\mathrm{lower}\mathrm{terms},$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}x+y.$$

Answered by Olaf last updated on 25/Dec/20

$$\mathrm{S}=\underset{n=0}{\sum ^{1008}}\frac{1}{(2n+1)(2n+3)}$$$$\mathrm{S}=\frac{1}{2}\underset{n=0}{\sum ^{1008}}[\frac{1}{2n+1}-\frac{1}{2n+3}]$$$$\mathrm{S}=\frac{1}{2}[1-\frac{1}{2019}]$$$$\mathrm{S}=\frac{1}{2}\times \frac{2018}{2019}=\frac{1009}{2019}$$$$x=1009,y=2019,x+y=3028$$

Commented by ZiYangLee last updated on 30/Dec/20

$$\mathrm{perfect}!$$

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