Question

1+3+5+...+99
2+4+6+...+100

Answer 1

Răspuns:

Explicație pas cu pas:

sume Gauss:

$\boxed{ 1 + 3+ 5 + ... + (2n - 1) = {n}^{2}}$

$1+3+5+...+99 = 1+3+5+...+(2 \cdot 50 - 1) =\bf 50^{2}$

$\boxed {1 + 2 + 3 + ... + n = \dfrac{n \cdot (n + 1)}{2}}$

$2+4+6+...+100 = 2 \cdot (1+2+3+...+50) = 2 \cdot \dfrac{50 \cdot 51}{2} = 50 \cdot 51 = \bf 2550$