Question

curbure
1+2+3+...+99/
2 +4 +6+…+100

Answer 1

folosim formula Sumei Gauss:

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

$\dfrac{1+2+3+...+99}{2+4+6+...+100} = \dfrac{1+2+3+...+99}{2 \cdot (1+2+3+...+50)} =$

$= \dfrac{\dfrac{99 \cdot 100}{2} }{2 \cdot \dfrac{50 \cdot 51}{2} } = \dfrac{99 \cdot 50 }{50 \cdot 51} = \dfrac{99^{(3} }{51} = \bf \dfrac{33}{17}$