Question

Calculati sumele:
a) 1+2+3+...100
b) 1+3+5+...+99
c) 2+4+6+...100

Answer 1

1+2+3+...+100=100*101:2=5050


1+3+5+...+99=1+3+...+(2*49+1)=(49+1)²=50²=2500

2+4+5+...+100=2*(1+2+3+...+50)=2*50*51:2=2550

Answer 2

$\displaystyle a).1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = \frac{10100}{2} =5050 \\ \\ b).1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2(1+2+3+...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} = 4950- \not2 \times \frac{2450}{\not 2} =4950-2450=2500$

$\displaystyle c).2+4+6+...+100=2(1+2+3+...+50)=2 \times \frac{50(50+1)}{2} = \\ \\ =2 \times \frac{50 \times 51}{2} =\not 2 \times \frac{2550}{\not 2} =2550$