Question 288282
Hint: Rewrite the series - 2 + 3 - 4 + 5 - 6 +. . . - 100 into (3+5+7+9+...97+99)+(-2-4-6-8-...-98-100). From there, break up 3 into 2*1+1, break up 5 into 2*2+1, break up 7 into 2*3+1, etc to get 2*1+1+2*2+1+2*3+1+...2*48+1+2*49+1. Also, rewrite 2, 4, 6, etc into 2*1, 2*2, 2*3, etc



So...


- 2 + 3 - 4 + 5 - 6 +. . . - 100 = (3+5+7+9+...97+99)+(-2-4-6-8-...-98-100) = (2*1+1+2*2+1+2*3+1+...2*48+1+2*49+1) + (-2*1-2*2-2*3-...-2*49-2*50)


Or simply 


- 2 + 3 - 4 + 5 - 6 +. . . - 100 = (2*1+1+2*2+1+2*3+1+...2*48+1+2*49+1) + (-2*1-2*2-2*3-...-2*49-2*50)



Now group all of the terms with a factor of 2 in (2*1+1+2*2+1+2*3+1+...2*48+1+2*49+1) to get ((2*1+2*2+2*3+...2*48+2*49)+1+1+1+...1+1)



Now we then get 


- 2 + 3 - 4 + 5 - 6 +. . . - 100 = ((2*1+2*2+2*3+...2*48+2*49)+1+1+1+...1+1) + (-2*1-2*2-2*3-...-2*49-2*50)



Factor out the '2's:


- 2 + 3 - 4 + 5 - 6 +. . . - 100 = (2(1+2+3+...48+49)+1+1+1+...1+1) + -2(1+2+3+...+49+50)



What we now have is a series 1+2+3+....+n in which we can use the formula {{{(n(n+1))/2}}}. I'll let you do that.