用裂项求和的办法计算下面各题:

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用裂项求和的办法计算下面各题:

用裂项求和的办法计算下面各题:
用裂项求和的办法计算下面各题:

用裂项求和的办法计算下面各题:
1、
原式=(1/1992-1/1993)+(1/1993-1/1994)+(1/1994-1/1995)+1/1995
=1/1992-1/1993+1/1993-1/1994+1/1994-1/1995+1/1995
=1/1992
2、
原式=(1+2+3+……+19+20)+(1/2+1/6+1/12+1/20+……+1/420)
=(1+20)x20÷2+[1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+……+1/(20x21)]
=210+[(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/20-1/21)]
=210+(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/20-1/21)
=210+1-1/21
=210又20/21
3、
原式=(1/2012+2011/2012)x1006÷2
=503
4、
原式=(1-1/2)+(1-1/4)+(1-1/8)+(1-1/16)+……+(1-1/1024)
=10-(1/2+1/4+1/8+1/16+……+1/1024)
=10-1/2x[1-(1/2)^10]/(1-1/2)
=10+(1-1/1024)
=10又1023/1024

(1)原式=(1/1992-1/1993)+(1/1993-1/1994)+(1/1994-1/1995)+1/1995=1/1992
(2)原式=[1/2+1/(2×3)+1/(3×4)+1/(4×5)+……+1/(20×21)]+(1+2+3+4+……+20)
=[1/2+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+……+(1/20-1/21)]+[20×(1+20)]/2
=(1/2+1/2-1/21)+210=210+20/21
=4430/21

(1)原理:1/(1992x1993)=1/1992-1/1993
如此类推,最后的结果是: 1/1992
(2)这题,我只会1又1/2=1+1+1/2,如此类推
然后分类求和,即1+2+3+4+.......+1/2+1/6+1/12+.......

(1)原式=1/1992-1/1993+1/1993-1/1994+1/1994-1/1995+1/1995
=1/1992
(2)原式=(1+2+3+4+。。。+20)+(1/2+1/6+1/12+1/20+...+1/420)
=210+(1-1/2+1/2-1/3+1/3-1/4+...+1/20-1/21)

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(1)原式=1/1992-1/1993+1/1993-1/1994+1/1994-1/1995+1/1995
=1/1992
(2)原式=(1+2+3+4+。。。+20)+(1/2+1/6+1/12+1/20+...+1/420)
=210+(1-1/2+1/2-1/3+1/3-1/4+...+1/20-1/21)
=210+(1-1/21)
=210+20/21
=210又21分之20
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收起

两题都要用到1/(n*(n+1))=1/n-1/(n+1)(1)原式=1/1992-1/1993+1/1993-1/1994+1/1994-1/1995+1/1995=1/1992 (2)原式=1+2+3+4+……+20+1/2+1/(2*3)+1/(3*4)+1/(4*5)+……+1/(20*21)=210+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/20-1/21=210+1-1/21=4430/21