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


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

$Let m and n be two positive integers$ $satisfy \frac{m}{n} = \frac{1}{10 \times 12} + \frac{1}{12 \times 14} + \frac{1}{14 \times 16} + . . . + \frac{1}{2012 \times 2014}$ $find the smallest possible value of$ $m + n$
	Answered by john santu last updated on 02/Jun/20

	$\Rightarrow \frac{m}{n} = \frac{1}{4} \underset{p = 5}{\sum^{1006}} \frac{1}{p (p + 1)}$ $= \frac{1}{4} \underset{p = 5}{\sum^{1006}} \frac{1}{p} - \frac{1}{p + 1} [telescopy]$ $= \frac{1}{4} (\frac{1}{5} - \frac{1}{1007}) = \frac{501}{10070}$ $since \gcd (501, 10070) = 1$ $we have m + n = 10571$
	Commented by selea last updated on 02/Jun/20

	$I think \frac{1}{4} has problem$
	Commented by john santu last updated on 02/Jun/20

	$why ?$
	Commented by bobhans last updated on 02/Jun/20

	$correct mr john . thanks$
	Commented by selea last updated on 02/Jun/20

	$\frac{1}{10 \times 12} + \frac{1}{12 \times 14} . . . + \frac{1}{2012 \times 2014} \neq \frac{1}{4} \underset{p = 5}{\sum^{1006}} (\frac{1}{p (p + 1)})$
	Commented by bobhans last updated on 02/Jun/20

	$\frac{1}{120} = \frac{1}{4 \times 5 \times 6} = \frac{1}{4} \times \frac{1}{5 \times 6}$ $\frac{1}{12 \times 14} = \frac{1}{4 \times 6 \times 7} = \frac{1}{4} \times \frac{1}{6 \times 7}$ $\frac{1}{14 \times 16} = \frac{1}{4 \times 7 \times 8} = \frac{1}{4} \times \frac{1}{7 \times 8}$
	Commented by bobhans last updated on 03/Jun/20

	$your wrong sir selea$
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