用简便方法计算 (1) 2004分之1+2004分之2+2004分之3+…+2004分之2003

问题描述:

用简便方法计算 (1) 2004分之1+2004分之2+2004分之3+…+2004分之2003
(2) 1+2又6分之1+3又12分之1+4又20分之1+5又30分之1+6又42分之1+7又56分之1+8又72分之1+9又90分之1
(3) 1×6分之1+6×11分之1+11×16分之1+…+101×106分之
(4) (1+0.12+0.23)+(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)

一题:
1/2004+2/2004+3/2004+……+2003/2004
=(1+2+3+…+2003)/2004
=【(1+2003)×2003÷2】/2004
=1001.5
二题:用分数的拆分法来做.
因为:1/6=1/2-1/3; 1/12=1/3-1/4; 1/20=1/4-1/5;
1/30=1/5-1/6; 1/42=1/6-1/7; 1/56=1/7-1/8; 1/72=1/8-1/9;
1+2又1/6+3又1/12+4又1/20+5又1/30+6又1/42+7又1/56+8又1/72+9又1/90
=(1+2+3+4+5+6+7+8+9)+(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=45+(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)
=45+(1/2-1/9)
=45又7/18
三题:(方法同上,用拆项法做)
因为,1/(1×6)=(1-1/6)×1/5;
1/(6×11)=(1/6-1/11)×1/5;
1/(11×15)=(1/11-1/15)×1/5;
…………………………………………
1/(101×106)=(1/101-1/105)×1/5;
1×6分之1+6×11分之1+11×16分之1+…+101×106分之
=(1-1/6)×1/5+(1/6-1/11)×1/5+(1/11-1/15)×1/5+……+(1/101-1/105)×1/5
=(1-1/6+1/6-1/11+1/11-1/15+……+1/101-1/105)×1/5
=(1-1/105)×1/5
=104/525
4题:(这道题用假设法来做,并且题目有错,正确的题目如下)
(1+0.12+0.23)×(0.12+0.23+0.34)-(1+0.12+0.23+0.34)×(0.12+0.23)
设:(1+0.12+0.23)=a; 0.12+0.23=b;
原式=a×(b+0.34)-(a+0.34)×b
=ab+0.34a-ab-0.34b
=0.34×(a-b)
=0.34×(1+0.12+0.23-0.12-0.23)
=0.34×1
=0.34