1+2-3+4+5-6+7+8-9+.+97+98-99+100=?

1+2-3+4+5-6+7+8-9+.+97+98-99+100=?

Every three items are divided into a group (with brackets) +96 + 100, a total of 34 items are added. The first 33 items are the arithmetic sequence with 3 as the tolerance and 0 as the first item, so the original formula = 33 (0 + 96) / 2 + 100 = 1684

(1-2 / 1) * (1-3 / 1) (1-4 / 1) *. * (1-99 / 1) (1-100 / 1)
Please write down the process, it should be simple!

(1/2)*(2/3)*(3/4).*(99/100)
The denominator of the former number can be eliminated from the numerator of the latter number, and only 1 / 100 is left

1 × (- 1 / 2) + 1 / 2 × (- 1 / 3) + 1 / 3 × (- 1 / 4) + + 1 / 99 × (- 1 / 100)
Using the method of sequence >

1 × (- 1 / 2) + 1 / 2 × (- 1 / 3) + 1 / 3 × (- 1 / 4) + + 1 / 99 × (- 1 / 100)
=-(1/1×2+1/2×3+1/3×4+…… +1/99×100)
=-(1/1-1/2+1/2-1/3+1/3+1/4-1/4+…… -1/99+1/99-1/100)
=-(1-1/100)
=-99/100