Question

aratati ca numarul A=radical (2016-1008/radical 1+3+5+7+...+2015)la putera 2015 :2015 este natural DAU COROANA LA CEL MAI BUN RASPUNS!!!!!

Answer 1

1+3+5+..+2015=
1=1+2*0
3=1+2*1
5=1+2*2
............
2015=1+1007*2
_____________________(+)
=1*1008 + 2(1+2+3+..+1007)=
=1008 + 2*1007*1008:2=
=1008 + 1007*1008
=1008(1+1007)
=1008*1008
=1008²

A = √ [(2016 - 1008/1008)²⁰¹⁵ : 2015]
A = √[(2016-1)²⁰¹⁵ : 2015]
A = √[2015 ²⁰¹⁵ : 2015]
A = √2015²⁰¹⁴
A = 2015¹⁰⁰⁷∈ N