2 2 2——— + ——— +···+ ——————— 1*2*3 2*3*4 2002*2003*20042/(1*

1个回答

  • 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)