Question

Aratati ca numarul
$n = 2019 + 2 \times (1 + 2 + 3 + ... + 2018)$
Este pătrat perfect ​

Answer 1

Răspuns:

Explicație pas cu pas:

1 + 2 + 3 + ....+ 2018 = 2018*(1 + 2018)/2 = 2018*2019/2

2*2018*2019/2 = 2018*2019

n = 2019 + 2018*2019 = 2019*(1 + 2018) = 2019*2019 = 2019^2 patrat perfect

Answer 2

Salutare!

n = 2019 + 2 × ( 1 + 2 + 3 + ... + 2018 )

Aplicam formula sumei lui GAUSS pentru paranteza.

1 + 2 + 3 + ... + 2018 = [ 2018 × (2018 + 1 ) ] / 2

1 + 2 + 3 + ... + 2018 = ( 2018 × 2019 ) / 2

n = 2019 + 2 × ( 1 + 2 + 3 + ... + 2018 )

n = 2019 + 2 × ( 2018 × 2019 ) / 2   → 2 cu 2 se simplifica

n =2019 + 2018 × 2019   → factor comun 2019

n = 2019 ( 1 + 2018 )

n = 2019 × 2019

n = 2019² → n patrat perfect