1/2*3*4+1/3*4*5....1/201*202*2

1/(2*3*4) = (1/2) * [1/(2*3) - 1/(3*4)]
1/(3*4*5) = (1/2) * [1/(3*4) - 1/(4*5)]
. . .
1/(2001*2002*2003) = (1/2) * [1/(2001*2002) - 1/(2002*2003)]
相加, 得：(sum up, get)
1/(2*3*4) + 1/(3*4*5) + ... + 1/(2001*2002*2003)
= (1/2) * [1/(2*3) - 1/(2002*2003)]
= (1/2) * (1001*2003 - 3)/(3*2002*2003)
= 2005000/(2*3*2002*2003)
= 501250/6015009

2012-09-25 10:45:01 補充：
公式 (formula)
1/[n(n + 1)(n + 2)] = (1/2) * {1/[n(n + 1)] - 1/[(n + 1)(n + 2)]}

∵ 1/[(n – 1)(n)(n + 1)]= ½[(n – 1) – (n + 1)]/[(n – 1)(n)(n + 1)]= ½/[(n)(n + 1)] - ½/[(n + 1)(n + 2)]= ½[(n + 1) – n]/[(n)(n + 1)]- ½[(n + 2) – (n + 1)]/[(n + 1)(n +2)]= ½/n - ½/(n + 1) - ½/(n + 1) + ½/(n + 2)= ½/n – 1/(n + 1) + ½/(n + 2) ∴ 1/(2*3*4) + 1/(3*4*5)+ …+ 1/(2001*2002*2003)= ½/2 - 1/3 + ½/4 + ½/3 – 1/4 + ½/5+ ½/4 – 1/5 + ½/6 + …+ ½/1999 – 1/2000 + ½/2001+ ½/2000 – 1/2001 + ½/2002+ ½/2001 – 1/2002 + ½/2003= ½/2 - ½/3 + 0 + … + 0 - ½/2002 + ½/2003= ¼ - ⅙ - 1/4004 + 1/4006=501250/6015009IHOPE THIS CAN HELP YOU! ~ ^_^