Question

Arătări ca numarul A = 4 * [(1+2+.....+2016)la puterea a 2 + 2017 la puterea a treia ] este produsul a doua pătrate perfecte consecutive . Va rog sa ma ajutati !

Answer 1

A = 4 * [(1+2+.....+2016)^2 +2017^3]

1+2+.....+2016={aplicam 1+2+...+n=[n(n+1)]/2 }=

(2016x2017)/2

A devine: 4*{[(2016x2017)/2]^2 +2017^3}=

=4*{[(2016x2017)^2/4] +2017^3}=

=[(2016x2017)^2] +4x2017^3=

=(factor comun 2017^2)=

=(2017^2)x[(2016^2)+4x2017]=

=(2017^2)x[(2016^2)+4x(2016+1)]=

=(2017^2)x[(2016^2)+4x2016+4]

In paranteza patrata avem un patrat perfect: a^2 +4a+4=(a+2)^2

La noi a=2016, asadar A devine:

A=(2017^2)x[(2016^2)+4x2016+4]=

=(2017^2)x[(2016+2)^2]=

(2017^2)x(2018^2)

A = produsul a doua pătrate perfecte consecutive.