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Answer 1 · 2018-01-22T19:13:24

#S=-2,037,171#

We can solve this using the difference of two squares.

#a^2-b^2=(a+b)(a-b)#

Using this fact we can write:

#S=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2017-2018)(2017+2018)#

#=-1(1+2)-1(3+4)-1(5+6)-...-1(2017+2018)#

Take out a factor of -1

#S=-1(1+2+3+4+...+2017+2018)#

This is -1 lots of the sum of the first 2018 integers.

#S=-sum_"r=1"^2018"r#

From standard results (#sum_"r=1"^n"r=1/2n(n+1)#)

#S=-2018/2xx(2018+1)#

#S=-2,037,171#