Răspuns:
q) (1+3+...+2001+2003)- (2+4+...2000+2002)
luam pe rand fiecare paranteza
(1+3+...+2001+2003)
1= 2x 0+1
3=2x 1+ 1
.
.
.
2001=2x1000 +1
2003= 2x1001 +1
de unde rezulta ca (1+3+...+2001+2003)= 2x(1+2+...+1001) +1x1001+1= 2x(1+2+...+1001)+ 1002
(2+4+...2000+2002)= 2x(1+2+...+1001)
(1+3+...+2001+2003)- (2+4+...2000+2002)=2x(1+2+...+1001)+ 1002-2x(1+2+...+1001)= 1002
