Question

Calculati sumele :
A) 1+2+3...+100
B)1+3+5+...+99
C)2+4+6+...+100
D)5+10+15+...+100

Answer 1

$\displaystyle a).1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = \frac{10100}{2} =5050 \\ \\ b).1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2(1+2+3+...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} =4950-\not2 \times \frac{2450}{\not2} =4950-2450=2500$

$\displaystyle c).2+4+6+...+100=2(1+2+3+...+50)=2 \times \frac{50(50+1)}{2} = \\ \\ =2 \times \frac{50 \times 51}{2} =\not2 \times \frac{2550}{\not2} =2550 \\ \\ d).5+10+15+...+100=5(1+2+3+...+20)=5 \times \frac{20(20+1)}{2} = \\ \\ =5 \times \frac{20 \times 21}{2} =5 \times \frac{420}{2} =5 \times 210=1050$