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| Question 288282:  The  value   of  - 2 + 3 - 4 + 5 - 6 +. . . - 100   is
 (a) -150 (b) =100 (c) -50 (d) -49 (e) 0
 
 Answer by jim_thompson5910(35256)
      (Show Source): 
You can put this solution on YOUR website! 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
  . I'll let you do that.
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