Explicație pas cu pas:
a)
[tex]\frac{1+2+ ... + 15}{2 + 4 + ... + 30} = \frac{1+2+ ... + 15}{2(1+2+ ... + 15)} = \frac{1}{2}[/tex]
b)
[tex]\frac{3 + 6 + 9 + ... + 120}{5 + 10 + 15 + ... + 200} = \frac{3(1 + 2 + ... + 40)}{5(1 + 2 + ... + 40)} = \frac{3}{5}[/tex]
c)
[tex]\frac{1 + 3 + 5 + ... + 101}{51} = \frac{(\frac{1 + 101}{2})^2}{51} = \frac{51^2}{51} = 51[/tex]
Suma nr impare de la 1 la 101 este: [tex](\frac{1 + 101}{2})^2 [/tex] conform formulii
