先找规律..
设每一个都是
2/(n-1)n(n+1) (n=2,3,4,5.2003)
=2/(n平方-1)n
=2*[n2-(n2-1)/(n2-1)n] (n2是n平方)
=2*(n/n2-1 - 1/n)
=2*(n+1/n2-1 - 1/n2-1 - 1/n)
=2*[ 1/n-1 - 1/n - 1/(n+1)(n-1) ]
=2*(1/n-1 - 1/n) - 2*[1/(n+1)(n-1)]
=2*(1/n-1 - 1/n) - 2* 1/2 * [1/(n-1) - 1/(n+1)]
把数字代进去就得到:
[2*(1- 1/2) - (1- 1/3)]+[2*(1/2 - 1/3) - (1/2 - 1/4)]+.+[2*(1/2002- 1/2003) - (1/2002 - 1/2004)]
= 2*(1-1/2 +1/2 - 1/3 + 1/3 - 1/4 +.+ 1/2001 -1/2002 + 1/2002 - 1/2003) - (1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 +.+ 1/2001 -1/2003 + 1/2002 - 1/2004)
=2*(1- 1/2003) - (1 + 1/2 - 1/2003 - 1/2004)
= -4011007/(2003*2004)