Question

dati nr 2 factor comun  2+4+6+...100

Answer 1

[tex]2+4+6+...100= \\ \\ 2(1+2+3+...+50)= \\ \\ 2 \frac{50(50+1)}{2} =50*51=2550[/tex]

Answer 2

2· ( 1+2+3+4+ ... + 50 )
2  · $\frac{n*(n+1)}{2}$
2 · $\frac{50*51}{2}$
( 2 cu 2 se simplifică ) ⇒ 50·51 = 2550.