a) $1+(-2)+3+(-4)+...+2017+(-2018)\\=(1+3+...+2017)-(2+4+...+2018)\\=2(\frac{1009^2}{2}+\frac{1009}{2})-1009-2(\frac{1009^2}{2}+\frac{1009}{2})\\=-1009$
b) $1+(-6)+11+(-16)+...+(-996)+1001\\=(1+11+...+1001)-(6+16+...+996)\\=10(\frac{101^2}{2}+\frac{101}{2})-9\times 101-10(\frac{100^2}{2}+\frac{100}{2})-4\times 100$
