Let S=1+2+3−4+5+6+7−8+⋯+197+198+199−200 We can rewrite it as: S=(1+2+3+⋯+200)−2(4+8+⋯+200)
The sum of the first 200 integers: 1+2+⋯+200=2200×201=20100 The second term is an arithmetic series of 4’s multiples: 4+8+⋯+200=4(1+2+⋯+50)=4(250×51)=4(1275)=5100 Therefore, S=20100−2(5100)=20100−10200=9900