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1/(2*3*4)+1/(3*4*5)+1/(4*5*6)+1/(5*6*7)+1/(6*7*...

1只+1只=1双; 1(个月)+2(个月)=1季度 3天+4天=1周 5个月+7个月=1年 6(小时)+18小时=1天

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...

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...

'非常简单Private Sub Command1_Click()Dim i As IntegerDim f, s As DoubleFor i = 2 To 19 If i Mod 2 = 0 Then s = (-1) * (1 / (i * (i + 1))) End If If i Mod 2 = 1 Then s = 1 / (i * (i + 1)) End If f = s + fNext if = 1 + fPrint "...

观察:2*4,3*5……,整个式子中只有它在变,设它为(n-1)*(n+1),因为这样的话,第一项取3,第二项取4…… 通分:得到每一项的通式(n^2-1-3)/(n^2-1)=(n+2)(n-2)/(n+1)(n-1) 【3

你把括号去掉,每个括号的最后一个和后一个括号的第一个相乘为1,然后就整理为1/2乘2015/2014等于2015/4028

解:1/(4×5×6)=1/2[1/(4×5)-1/(5×6)] 1/(5×6×7)=1/2[1/(5×6)- 1/(6×7 )] 1/(6×7×8)=1/2[1/(6×7)- 1/(7×8)] ……… 1/(98×99×100)=1/2[1/(98×99)-1/(99×100)] ∴1/(4×5×6)+ 1/(5×6×7)+ 1/(6×7×8)+…..+ 1/(98×99×100) =1/2[1/(4×5)-1/(5×6)]+ 1/2[1/(5...

1/4+3/4*1/3 =¼×(1+3×1/3) =¼×(1+1) =¼×2 =½

下面是你的代码修改后并能成功运行的代码 #include main() { float s,a=1.0,t; s=0; int i; for(i=1;i

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*...

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