數學問題[十萬火急]!

1/6 = 1/(2 x 3)(-1)^0
-1/12 = 1/(3 x 4)(-1)^1
1/20 = 1/(4 x 5)(-1)^2
-1/30 = 1/(5 x 6)(-1)^3
therefore the nth term is (-1)^(n-1)/[(n + 1)(n + 2)]

分母相差6、8、10……

1st term = 1/6 = 1/(2*3)
2nd term = -1/12 = -1/(3*4)
3rd term = 1/20 = 1/(4*5)
4th term = -1/30 = -1(/5*6)

therefore nth term = (-1)^n/[(n+1)(n+2)]