Question

A=2019+2×(1+2+3+...+2018)
B=1+3+5+...+2019
Să se arate că:
a)A şi B sunt pătrate perfecte
b) Arătați că nr. C=(2019×2019×2019)-(2019×2019×2018)-(2019×2018)+6 este pătrat perfect

Answer 1

a) A = 2019 + 2*(2018 * 2019 : 2) = 2019 + 2*(1008 * 2019) = 2019 + 2018 * 2019 = 2019 * (1 + 2018) = 2019 * 2019 = 2019^2 ⇒ A este patrat perfect

B = 1 + 3 + 5 +...+ 2019 = 1 + 1 + 2*1 + 1 + 2*2 + ... + 1 + 2*1008 = 1009 + 2 * (1 + 2 + 3 + 4 +...+ 1008) = 1009 + 2 * (1008 * 1009 : 2) = 1009 + 1008 * 1009 = 1009 (1 + 1008) = 1009 * 1009 = 1009^2 ⇒ B este patrat perfect

b) C = 2019 * 2019 * 2019 - 2019 * 2019 * 2018 - 2019 * 2018 + 6 =

= 2019 * 2019 * (2019 - 2018) - 2019 * 2018 + 6 = 2019 * 2019 - 2019 * 2018 + 6 = 2019 * (2019 - 2018) + 6 = 2019 + 6 = 2025 = 45^2 ⇒ C este patrat perfect

Answer 2

$A = 2019+2\cdot(1+2+3+...+2018) = 2019+2\cdot \dfrac{2018\cdot 2019}{2} = \\ = 2019+2018\cdot 2019 = 2019\cdot (1+2018) = 2019\cdot 2019 = 2019^2 \\ \\ B = 1+3+5+...+2019 = \left(\dfrac{2019+1}{2}\right)^2 = \left(\dfrac{2020}{2}\right)^2 = 1010^2\\ \\ C= (2019\cdot 2019\cdot 2019)-(2019\cdot 2019\cdot 2018)-(2019\cdot 2016)+6 = \\ \\ =2019^3-2019^2\cdot 2018-2019\cdot 2018+6 = \\ = 2019^2\cdot (2019-2018) - 2019 \cdot 2018+6 = \\ =2019^2\cdot 1 - 2019\cdot 2016+6 = \\ = 2019\cdot (2019-2018)+6 =$

$= 2019\cdot 1+6 = 2019+6 = 2025 = 45^2$