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

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

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

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

⇒ (m/n) = (1/4)Σ_(p = 5) ^(1006) (1/(p(p+1))) = (1/4)Σ_(p = 5) ^(1006) (1/p)−(1/(p+1)) [ telescopy] =(1/4)((1/5)−(1/(1007))) = ((501)/(10070)) since gcd(501,10070) = 1 we have m+n = 10571

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

Commented by selea last updated on 02/Jun/20

I think (1/4) has problem

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

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

why?

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

Commented by bobhans last updated on 02/Jun/20

correct mr john. thanks

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

Commented by selea last updated on 02/Jun/20

(1/(10×12))+(1/(12×14))...+(1/(2012×2014))≠(1/4)Σ_(p=5) ^(1006) ((1/(p(p+1))))

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

Commented by bobhans last updated on 02/Jun/20

(1/(120))= (1/(4×5×6)) = (1/4)×(1/(5×6)) (1/(12×14))=(1/(4×6×7))=(1/4)×(1/(6×7)) (1/(14×16)) = (1/(4×7×8))=(1/4)×(1/(7×8))

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

Commented by bobhans last updated on 03/Jun/20

your wrong sir selea

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

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