Question

calculati următoarele sume:b)2+4+6+....+100 c)1+3+5+....+99

Answer 1

$\displaystyle a).2+4+6+...+100=2(1+2+3+...+50)=2 \times \frac{50(50+1)}{2} = \\ \\ =2 \times \frac{50 \times 51}{2} =\not2 \times \frac{2550}{\not2} =2550 \\ \\ b).1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2(1+2+3+...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} =4950- \not2 \times \frac{2450}{\not 2} =4950-2450=2500$