[tex]$ \ $ Folosim formulele: \\ \\ $ 1+3+5+...+x = (\frac{x+1}{2})^2 \\ \\ 1+2+3+...+x = \frac{x(x+1)}{2} \\ \\ \\ S_1 = 1-2+3-4+5-...-98+99-100 \\ S_1 = 1+3+5+...+99 -2-4-6-...-100 \\ S_1 = 1+3+5+...+99-2(1+2+3+...+50) \\ \\ \RIghtarrow S_1 = (\frac{99+1}{2})^2 - 2\cdot \frac{50(50+1)}{2} \Rightarrow S_1 = (\frac{100}{2})^2 - 2\cdot \frac{50\cdot 51}{2} \Rightarrow \\ \\ \RIghtarrow S_1 = 50^2 - 50\cdot 51 \Rightarrow S_1 = 50(50-51) \Rightarrow S_1 = 50\cdot( -1) \Rightarrow \\ [/tex]
[tex]\Rightarrow S_1 = -50 \\ \\ S_2 = -2-4-6-...-100 \Rightarrow S_2 = -2(1+2+3+...+50)\Rightarrow \\ \\ \Rightarrow S_2 = -2\cdot \frac{50\cdot51}{2} \Rightarrow S_2 = -50 \cdot 51 \Rightarrow S_2 = -2550[/tex]
