p-1-1-2-1-3-1-4-1-2003-1-2004-q-1-1003-1-1004-1-2004-p-2-q-2-

Tinku Tara June 4, 2023 Algebra 0 Comments

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

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

$${p}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2003}}−\frac{\mathrm{1}}{\mathrm{2004}} \\ $$$${q}=\frac{\mathrm{1}}{\mathrm{1003}}+\frac{\mathrm{1}}{\mathrm{1004}}+…+\frac{\mathrm{1}}{\mathrm{2004}} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} \:=\: \\ $$

Answered by naka3546 last updated on 30/Apr/20