Question

先观察每个算式的特点，再巧算

1. 1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90

2. 1/5*7+1/7*9+1/9*11+1/11*13+1/13*15

3. 1/1*2+1/2*3+1/3*4+…+1/2010*2011

4. 1/3+1/15+1/35+1/63+…+1/17*19

5. (1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)

Answer 1

观察1、每个分母为1X2、2X3、3X4、4X5、5X6、6X7、7X8、8X9、9X10，而
1/2=1-1/2
1/6=1/2-1/3
1/12=1/3-1/4
1/20=1/4-1/5
1/30=1/5-1/6
......
1/90=1/9-1/10
∴ 1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10
=1-1/10
=9/10
此类型题目为分式裂项拆分法,消除中间项，解答较为简便。同理
2、 1/5*7+1/7*9+1/9*11+1/11*13+1/13*15
=1/2[1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15]
=1/2[1/5-1/15]
=1/2[3/15-1/15]
=1/2X(2/15)
=1/15
3、1/1*2+1/2*3+1/3*4+…+1/2010*2011
=1-1/2+1/2-1/3+...+1/2010-1/2011
=1-1/2011
=2010/2011
4、1/3+1/15+1/35+1/63+…+1/17*19
=1/3+1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/17-1/19)
=1/3+1/2(1/3-1/19)
=1/3+1/6-1/38
=1/2-1/38
=19/38-1/38
=18/38
=9/19
5、 (1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)
= (1+1/2+1/3+1/4)*（【1/2+1/3+1/4】+1/5)-(【1+1/2+1/3+1/4】+1/5)*(1/2+1/3+1/4)
= (1+1/2+1/3+1/4)*(1/2+1/3+1/4)+1/5* (1+1/2+1/3+1/4)-(1+1/2+1/3+1/4)* (1/2+1/3+1/4)
-1/5*(1/2+1/3+1/4)
=1/5*(1+1/2+1/3+1/4)-1/5*(1/2+1/3+1/4)
=1/5+1/5*(1/2+1/3+1/4)-1/5*(1/2+1/3+1/4)
=1/5

追问

请问第二道题的1/2是什么？

追答

∵ [1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15]=2/35+2/63+2/99+...+2/195
也就是上面拆开项再合并时，因为分母中2个数字之差均为2，且在分子上，所以把2当成公因式提出来，但是和原来的式子之和均大了2倍，所以再乘1/2，这样式子就还原为原来等式了，这种式子变形在数学中经常会用到。

Answer 2

1、原式=1-1/2+1/2-1/3+1/3-1/4+……+1/8-1/9+1/9-1/10=1-1/10=9/10
2、原式=1/2×（1/5-1/7+1/7-1/9+……+1/13-1/15）=1/2×（1/5-1/15)=1/15
3、原式=1-1/2+1/2-1/3+……+1/2010-1/2011=1-1/2011=2010/2011
4、原式=1/2×（1-1/3+1/3-1/5+……+117-1/19)=1/2×（1-1/19)=9/19
5、原式=1×（1/2+1/3+1/4+1/5）+(1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)
-1×(1/2+1/3+1/4)-（1/2+1/3+1/4+1/5）×(1/2+1/3+1/4)
=1/2+1/3+1/4+1/5-(1/2+1/3+1/4)=1/5