Question

一道初中数学计算题:1/1*3+1/2*4+1/3*5+1/4*6+1/5*7+1/6*8------1/17*19+1/18*20 怎样做

Answer 1

1/1*3+1/2*4+1/3*5+1/4*6+1/5*7+1/6*8------1/17*19+1/18*20
观察知道，这些分数可以分组进行裂项相消计算。
原式=[/1*3+1/3*5+1/5*7+---+1/17*19]+
[1/2*4+1/4*6+1/6*8+---+1/18*20]
=1/2[1-1/3+1/3-1/5+---+1/17-1/19]
+1/2[1/2+1/4+1/4-1/6+---1/18-1/20]
=1/2[1-1/19]+1/2[1/2-1/20]
=9/19+9/40
=531/760

Answer 2

可以分解 原式可以化简为
1/2(1-1/3)+(1/2-1/4)+1/2(1/3-1/5)+1/4-1/6+1/2(1/5-1/7)+`````1/2(1/17-1/19)+（1/18-1/20）
即
1/2【(1-1/3+1/3-1/5+1/5-1/7+1/7-```1/17-1/19)】+【(1/2-1/4+1/4-1/6+1/6``````+1/18-1/20)】
1/2【(1-1/19)】+【(1/2-1/20)】=9/19+9/20=351/380

Answer 3

1/1*3 +1/2*4 +1/3*5 +1/4*6 +......+1/17*19 +1/18*20
=(1/1*3 +1/3*5+......+1/17*19 )+(1/2*4 +1/4*6 +......+1/18*20 )
=1/2 *[(1 - 1/3)+(1/3 - 1/5)+......+(1/17 - 1/19)] + 1/2 *[(1/2 - 1/4)+(1/4 - 1/6)+......+(1/18 - 1/20)]
=1/2*[(1-1/3+1/3-1/5+......+1/17-1/19)+(1/2-1/4+1/4-1/6+......+1/18-1/20)]
=1/2*[(1-1/19)+(1/2-1/20)]
=1/2*(18/19 +9/20)
=1/2* (521/380)
=521/760

这道题利用了两次分数拆分的方法。
公式：1/n(n+d)=1/d *[1/n - 1/n+d]

Answer 4

此题根据裂项求和公示计算，具体步骤为：
原式=1/2(1-1/3)+1/2（1/2-1/4)+1/2(1/3-1/5)+1/2(1/4-1/6)+.....+1/2(1/17-1/19)+1/2(1/18-1/20）
=1/2(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.......1/17-1/19+1/18-1/20)
=1/2(1+1/2-1/19-1/20)
=1/2*531/380
=531/760

Answer 5

原式=1/2(1-1/3)+1/2（1/2-1/4)+1/2(1/3-1/5)+1/2(1/4-1/6)+.....+1/2(1/17-1/19)+1/2(1/18-1/20）
=1/2(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6+.......1/17-1/19+1/18-1/20)
=1/2(1+1/2-1/19-1/20)
=1/2*531/380
=531/760