Question Number 126963 by ZiYangLee last updated on 25/Dec/20
$$\mathrm{If}\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}×\mathrm{9}}+\ldots+\frac{\mathrm{1}}{\mathrm{2017}×\mathrm{2019}}=\frac{{x}}{{y}} \\ $$$$\mathrm{where}\:\frac{{x}}{{y}}\:\mathrm{is}\:\mathrm{in}\:\mathrm{its}\:\mathrm{lower}\:\mathrm{terms},\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}+{y}. \\ $$
Answered by Olaf last updated on 25/Dec/20
![S = Σ_(n=0) ^(1008) (1/((2n+1)(2n+3))) S = (1/2)Σ_(n=0) ^(1008) [(1/(2n+1))−(1/(2n+3))] S = (1/2)[1−(1/(2019))] S = (1/2)×((2018)/(2019)) = ((1009)/(2019)) x = 1009, y = 2019, x+y = 3028](https://www.tinkutara.com/question/Q126964.png)
$$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{1008}} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\mathrm{1008}} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}\right] \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2019}}\right] \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2018}}{\mathrm{2019}}\:=\:\frac{\mathrm{1009}}{\mathrm{2019}} \\ $$$${x}\:=\:\mathrm{1009},\:{y}\:=\:\mathrm{2019},\:{x}+{y}\:=\:\mathrm{3028} \\ $$
Commented by ZiYangLee last updated on 30/Dec/20
$$\mathrm{perfect}! \\ $$