第1个回答 2008-10-05
1/[n*(n+1)(n+2)]=[1/n(n+1)-1/(n+1)(n+2)]/2
原式=[(1/2*3-1/3*4)+(1/3*4-1/4*5)+……+1/n(n+1)-1/(n+1)(n+2)]/2
=[1/6-1/(n+1)(n+2)]/2
第2个回答 2008-10-05
1/(2*3*4)+1/(3*4*5)+……+1/(98*99*100)
1/(2*3*4)=1/2-1/3-1/3+1/4
1/(3*4*5)=1/3-1/4-1/4+1/5
所以 原始=(1/2-1/3-1/3+1/4)+(1/3-1/4-1/4+1/5)+……+(1/98-1/99-1/99+1/100)
=1/2-1/3-1/99+1/100
第3个回答 2008-10-05
把中间项提出来 1/(2*3*4) = 1/3 * 1/(2*4)
= 1/3 * 1/2 * (1/2 -1/4)
= 2/3 * (1/2-1/4)
原式 = 2/3 * (1/2-1/4)+ 2/4 * (1/3 -1/5)+2/6 * (1/5-1/7)+
…… + 2/99 *(1/98 - 1/100)
展开 = 2/3* 1/2 -2/3 * 1/4 + 2/4 *1/3 -2/4 * 1/5 + 2/6 * 1/5 -
…… +2/99 * 1/98 - 2/99 * 1/100
中间项相抵消 =2/3 * 1/2 - 2/99 * 1/100
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