Let-m-and-n-be-two-positive-integers-satisfy-m-n-1-10-12-1-12-14-1-14-16-1-2012-2014-find-the-smallest-possible-value-of-m-n-

Question Number 96558 by bobhans last updated on 02/Jun/20

$$\mathrm{Let}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n}\:\mathrm{be}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{integers}\: \\ $$$$\mathrm{satisfy}\:\frac{\mathrm{m}}{\mathrm{n}}\:=\:\frac{\mathrm{1}}{\mathrm{10}×\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{12}×\mathrm{14}}+\frac{\mathrm{1}}{\mathrm{14}×\mathrm{16}}+…+\frac{\mathrm{1}}{\mathrm{2012}×\mathrm{2014}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{m}+\mathrm{n}\: \\ $$

Answered by john santu last updated on 02/Jun/20

$$\Rightarrow\:\frac{\mathrm{m}}{\mathrm{n}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{p}\:=\:\mathrm{5}} {\overset{\mathrm{1006}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{p}\:=\:\mathrm{5}} {\overset{\mathrm{1006}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{p}}−\frac{\mathrm{1}}{\mathrm{p}+\mathrm{1}}\:\:\left[\:\mathrm{telescopy}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{1007}}\right)\:=\:\frac{\mathrm{501}}{\mathrm{10070}} \\ $$$$\mathrm{since}\:\mathrm{gcd}\left(\mathrm{501},\mathrm{10070}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{m}+\mathrm{n}\:=\:\mathrm{10571} \\ $$

Commented by selea last updated on 02/Jun/20

$$\mathrm{I}\:\mathrm{think}\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{has}\:\mathrm{problem} \\ $$

Commented by john santu last updated on 02/Jun/20

$$\mathrm{why}? \\ $$

Commented by bobhans last updated on 02/Jun/20

$$\mathrm{correct}\:\mathrm{mr}\:\mathrm{john}.\:\mathrm{thanks} \\ $$

Commented by selea last updated on 02/Jun/20

$$\frac{\mathrm{1}}{\mathrm{10}×\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{12}×\mathrm{14}}…+\frac{\mathrm{1}}{\mathrm{2012}×\mathrm{2014}}\neq\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{p}=\mathrm{5}} {\overset{\mathrm{1006}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)}\right) \\ $$

Commented by bobhans last updated on 02/Jun/20

$$\frac{\mathrm{1}}{\mathrm{120}}=\:\frac{\mathrm{1}}{\mathrm{4}×\mathrm{5}×\mathrm{6}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{5}×\mathrm{6}} \\ $$$$\frac{\mathrm{1}}{\mathrm{12}×\mathrm{14}}=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{6}×\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{6}×\mathrm{7}} \\ $$$$\frac{\mathrm{1}}{\mathrm{14}×\mathrm{16}}\:=\:\frac{\mathrm{1}}{\mathrm{4}×\mathrm{7}×\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}}{\mathrm{7}×\mathrm{8}} \\ $$

Commented by bobhans last updated on 03/Jun/20

$$\mathrm{your}\:\mathrm{wrong}\:\mathrm{sir}\:\mathrm{selea} \\ $$

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