小学数学题目 1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4+5).......1/(1+2+3+4+....99)

小学六年级的题,请高手用小学生看得懂的方法计算,谢谢了

1+2=2*3/2
1+2+3=3*4/2
1+2+3+4=4*5/2
1+2+3+4+5=5*6/2
..........
1+2+3+4+......99=99*100/2
所以
1/(1+2)=2/(2*3)=2/2-2/3
1/(1+2+3)=2/(3*4)=2/3-2/4
1/(1+2+3+4)=2/(4*5)=2/4-2/5
1/(1+2+3+4+5)=2/(5*6)=2/5-2/6
..........
1/(1+2+3+4+......99)=2/(99*100)=2/99-2/100
所以
结果为2/2-2/100=49/50
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第1个回答  2010-06-28
1+2+3+----+n=n(n+1)/2
1/(1+2+3+---+n)=2/n(n+1)=2*[1/n-1/(n+1)]

1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4+5).......1/(1+2+3+4+....99)
=2*(1/2-1/3+1/3-1/4+1/4-1/5+......+1/99-1/100)
=2*(1-1/100)
=2*99/100
=99/50
第2个回答  2010-06-28
因为1+2+3+...+n=n(n+1)/2

因此 1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+...+99)
=2/(2*3)+2/(3*4)+2/(4*5)+...+2/(99*100)

又因为1/n(n+1)=1/n-1/(n+1)

因此 2/(2*3)+2/(3*4)+2/(4*5)+...+2/(99*100)
=(2/2-2/3)+(2/3-2/4)+(2/4-2/5)+...+(2/99-2/100)
=1-2/100
=49/50
第3个回答  2010-06-28
1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+1/(1+2+3+4+5).......1/(1+2+3+4+....99)
=2*(1/3+1/6+1/10+...+1/4950)/2
=2*(1/2*3+1/3*4+1/4*5+...+1/99*100)
=2*(1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100)
=2*(1/2-1/100)
=1-1/50
=49/50.
分析:利用1/6=1/2-1/3,1/12=1/3-1/4,即1/a(a+1)=1/a-1/(a+1)这种技巧
第4个回答  2010-06-28
49/50
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