p=1−(1/2)+(1/3)−(1/4)+...+(1/(2003))−(1/(2004)) q=(1/(1003))+(1/(1004))+...+(1/(2004)) p^2 +q^2 =


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Question Number 91277 by john santu last updated on 29/Apr/20

$p = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{2003} - \frac{1}{2004}$ $q = \frac{1}{1003} + \frac{1}{1004} + . . . + \frac{1}{2004}$ $p^{2} + q^{2} =$
	Answered by naka3546 last updated on 30/Apr/20

	Commented by jagoll last updated on 30/Apr/20

	$i t h i n k i t w r o n g a n s w e r$
	Commented by naka3546 last updated on 30/Apr/20

	$S h o w y o u r w o r k i n g s, p l e a s e$
	Answered by naka3546 last updated on 30/Apr/20

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