Question

这些算式有什么规律吗，再仿写一个

1/(1×2) + 1/(2×3) + 1/(3×4) + ... + 1/(2007×2008)
1/(1×3) + 1/(3×5) + 1/(5×7) + ... + 1/(2009×2011)
1/(2×4) + 1/(4×6) + 1/(6×8) + ... + 1/(2010×2012)

Answer 1

1/(1×2) + 1/(2×3) + 1/(3×4) + ... + 1/(2007×2008)
=1-1/2008=2007/2008
1/(1×3) + 1/(3×5) + 1/(5×7) + ... + 1/(2009×2011)
=1/2*(1-1/2011)
=1/2*2010/2011
=1005/2011
1/(2×4) + 1/(4×6) + 1/(6×8) + ... + 1/(2010×2012)
=1/2*(1/2-1/2012)
=1/2*2005/2012
=2005/4024

Answer 2

1/(1×2) + 1/(2×3) + 1/(3×4) + ... + 1/(2007×2008)
=1-1/2008=2007/2008
1/(1×3) + 1/(3×5) + 1/(5×7) + ... + 1/(2009×2011)
=1/2*(1-1/2011)
=1/2*2010/2011
=1005/2011
1/(2×4) + 1/(4×6) + 1/(6×8) + ... + 1/(2010×2012)
=1/2*(1/2-1/2012)
=1/2*2005/2012
=2005/4024
1/(1×4) + 1/(4×7) + 1/(7×10) + ... + 1/(2008×2011)
=1/3*(1-1/2011)
=1/3*2010/2011
=670/2011
内部可以抵消！

追问

1/(2×4) + 1/(4×6) + 1/(6×8) + ... + 1/(2010×2012)
=1/2*(1/2-1/2012)
=1/2*2005/2012
=2005/4024
这个我要具体过程，我还是个小学生呢，帮一下吧

追答

1/(2×4) + 1/(4×6) + 1/(6×8) + ... + 1/(2010×2012)
=1/2*(1/2-1/4+1/4-1/6+1/6-1/8+.....+1/2010-1/2012)
=1/2*(1/2-1/2012)
=1/2*2005/2012
=2005/4024

追问

规律到底是什么啊？看不懂，写详细一点吧，谢谢你