Question

1) a) 1+2+3+...+100=
b) 2+4+6+...+100=
c) 1+3+5+...+99=

2)2004+2005ori2004-2006ori2003=

Answer 1

$\displaystyle a).1+2+3+...+100= \frac{100(100+1)}{2} = \frac{100 \times 101}{2} = \frac{10100}{2} =5050 \\ \\ b).2+4+6+...+100=2(1+2+3+...+50)=2 \times \frac{50(50+1)}{2} = \\ \\ =2 \times \frac{50 \times 51}{2} =\not 2 \times \frac{2550}{\not 2} =2550$

$\displaystyle c).1+3+5+...+99= \\ \\ =1+2+3+4+5+...+99-(2+4+6+...+98)= \\ \\ = \frac{99(99+1)}{2} -2(1+2+3+...+49)= \frac{99 \times 100}{2} -2 \times \frac{49(49+1)}{2} = \\ \\ = \frac{9900}{2} -2 \times \frac{49 \times 50}{2} =4950 - \not 2 \times \frac{2450}{\not 2} =4950-2450=2500 \\ \\ d).2004+2005 \times 2004-2006 \times 2003= \\ \\ =2004+4018020-4018018= \\ \\ =4020024-4018018= \\ \\ =2006$