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1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7+1/7*8+1/8*9+1

拆分每个分数。记住等式: 1/(x(x+1))=1/x-1/(x+1) 所以,原等式就可以简化为1/1-1/10=0.9

1、这是一道六年级的数奥题,要用“分数的拆分法”来解答才行。 2、就是说,把题目中的所有分数拆成两个分数相减的形式,形成加减相互抵消,从而达到不同分的目的。 3、这道题首先把题中的分数拆分成下面的样子: 1/1*2*3=[1/(1×2)-1/(2×3)]×...

1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+1/5*6*7+1/6*7*8+1/7*8*9+1/8*9*10+1/9*10*11 =(1/2)[(1/1*2)-(1/2*3)]+(1/2)[(1/2*3)-(1/3*4)]+(1/2)[(1/3*4)-(1/4*5)]+...+(1/2)[(1/9*10)-(1/10*11)] =(1/2)[(1/1*2)-(1/2*3)+(1/2*3)-(1/3*4)+(1/3*4)-(1/4*...

原是=4/2(1/1*2-1/2*3)+5/2(1/2*3-1/3*4)+6/2(1/3*4-1/4*5)+…… +11/2(1/8*9-1/9*10) 这是第一次裂项,将(1*2*3)分之4裂成4/2(1/1*2-1/2*3) (2*3*4)分之5裂成5/2(1/2*3-1/3*4)……依次类推 继续运算得 原是=1+1/2{1/2*3+1/3*4+……+1/8*9}-(11/2...

1/(1×2)=(1/1)-(1/2) 1/(2×3)=(1/2)-(1/3) 1/(3×4)=(1/3)-(1/4) 1/(4×5)=(1/4)-(1/5) …… 1/(7×8)=(1/7)-(1/8) 则这个式子的值=(1/1)-(1/8)=7/8

1/[n(n+1)(n+2)]=1/2*[1/n-2/(n+1)+1/(n+2)], 例如 ∴1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+1/(4*5*6)+1/(5*6*7)+1/(6*7*8)+1/(7*8*9) =1/2*[1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+……+1/7-2/8+1/9] =1/2[1-1/2-1/8+1/9] =1/2(1/2-1/72) =35/144. 后面加一...

=1009/2019

1/((n-1)*n*(n+1))=0.5*(1/(n-1)*n-1/n*(n+1))=0.5*[(1/(n-1)-1/n)-(1/n-1/(n+1))]=0.5*(1/(n-1)+1/(n+1)-2/n) 也就是说1乘2乘3的分之一就是1加上3分之一减去2分之2,再乘以0.5后得到结果6分之一 记住这个公式哦 高三毕业都要用的哈 努力啊

解: 1/(1×3)+1/(2×4)+1/(3×5)+1/(4×6)+……+1/(99×101) =[1/(1×3)+1/(3×5)+…+1/(99×101)]+[1(2×4)+1/(4×6)+1/(6×10)+…+1/(96×98)] =(1-1/3+1/3-1/5+…+1/99-1/101)/2+(1/2-1/4+1/4-1/6+…+1/96-1/98)/2 =(1-101)/2+(1/2-1/98)/2 =50/101+12/49 =366...

A=1+1*2+1*2*3+1*2*3*4+1*2*3*4*5+1*2*3*4*5*6 =(1*2*3*4+1)+(1*2*3*4*5+1)+(1*2*3*4*5*6+1)+1*2*3 =5*5+11*11+19*19+6 =513 1*2*3*4+1=25=5*5 2*3*4*5+1=121=11*11 3*4*5*6+1=361=19*19 a(a+1)(a+2)(a+3)+1=(a²+3a+1)的平方...

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