谁能解这道数学题:1⼀(1*2*3)+1⼀(2*3*4)+1⼀(3*4*5)+......1⼀(26*27*28)=?

2024-12-17 20:20:56
推荐回答(4个)
回答1:

1/(1*2*3)=1/2*[1/1*2-1/2*3]=1/2*[(1/1-1/2)-(1/2-1/3)]=1/2*[1/1-2/2+1/3]
1/(2*3*4)=1/2*[1/2*3-1/3*4]=1/2*[(1/2-1/3)-(1/3-1/4)]=1/2*[1/2-2/3+1/4]
1/(3*4*5)=1/2*[1/3*4-1/4*5]=1/2*[(1/3-1/4)-(1/4-1/5)]=1/2*[1/3-2/4+1/5]
..............................................................................
1/(26*27*28)=1/2*[1/26*27-1/27*28]=1/2*[(1/26-1/27)-(1/27-1/28)]=1/2*[1/26-2/27+1/28]
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+.....+1/(26*27*28)
=1/2*{[1/1-2/2+1/3]+*[1/2-2/3+1/4]+[1/3-2/4+1/5]+...+[1/26-2/27+1/28]}
=1/2*{1-1/2-1/27+1/28}
=1/2*377/756
=377/1512

回答2:

1/(1*2*3)=[1/(1*2)-1/(2*3)]除以2 依次类推,可以一一消去

回答3:

1/【n*(n+1)*(n+2)】=【1/n+1/(n+2)-2/(n+1)】/2
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+......1/(26*27*28)
=【1/1+1/3-2/2+1/2+1/4-2/3+1/3+1/5-2/4+……+1/26+1/28-2/27】/2
=【1/1-1/2-1/27+1/28】/2
=377/1512

回答4:

解:因为:1/1-1/2-1/3=1/(1*2*3)
所以:原式=(1/1-1/2-1/3)+(1/2-1/3-1/4)+(1/3-1/4-1/5)+......+(1/26-1/27-1/28)
=1-1/2-1/3+1/3+1/2-1/3-1/4+1/3-1/4-1/5+......+1/26-1/27-1/28
=1-1/3-1/4-1/5-......-1/26-1/27-1/28
=1/(1*3*4*5*......*26*27*28) (这样写就可以了,因为乘出的数太大了)
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