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7分之5*11分之6*7/10 3/4*(9/10%5/6) (1/5+1/9...

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...

(9分之8+7分之10+11分之6)÷(11分之3+9分之4+7分之5) =2x(9分之4+7分之5+11分之3)÷(9分之4+7分之5+11分之3) =2。

(9分之8+1又7分之3+10分之6)除以(11分之3+7分之5+9分之4) =(8/9+10/7+6/10)除以(3/11+5/7+4/9) =(560/630+900/630+378/630)除以(189/693+495/693+308/693) =(1838/630)除以(993/693) =(1838/630)X(693/993) =(919/105)X(...

考察一般项第k项: (2k+1)/(1+2+...+k)²=(2k+1)/[k(k+1)/2]²=4[1/k² -1/(k+1)²] 3/1²+ 5/3²+ 7/6²+ 9/10² +11/15² =4(1/1²-1/2²+1/2²-1/3²+1/3²-1/4²+1/4²-1/...

分数七分之三八分之五九分之五中最接近二分之一的是()。分数4/5十分之九七分之六中最接近一的是() 1/2-3/7 =7/14-6/14 =1/14 5/8-1/2 =5/8-4/8 =1/8 5/9-1/2 =10/18-9/18 =1/18 最小的是1/18 所以 最接近2分之1的是9分之5. 1-4/5=1/5 1-9/10...

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*...

=(1-1/2)+(1-1/3)+...(1-1/10) =9-(1/2+1/3+...1/10) =9-[1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10] =9-[(1/2+1/3+1/6)+(1/4+1/5+1/10)+1/7+1/8+1/9] =9-(1+11/20+1/8+1/7+1/9) =9-(1+27/40+1/9+1/7) =9-(1+283/360+1/7) =8-2341/2520

将四个数字看成一组 (1+2-3-4)+(5+6-7-8)+……+(2005+2006-2007-2008) =(-4)+(-4)+…+(-4) =(-4)*(2008/4) =-2008 这里2008/4是因为一共有2008个数,每四个数一组,当然就是2008/4组数了,每组都是-4,两者相乘即可

首先提取公因式,归纳一下,就可以化成1/15*(1+2+.....+14)+1/14*(1+2+....+13)+........1/3*(1+2)+1/2 然后根据括号内可以用等差求和的方法每个单项式归纳为1/n*n/2*(n-1)=(n-1)/2 所以将N分别取15到2 提取公因式1/2 最后变成(1+3+4+......

解:首先找出an的通项公式。 an=(5n^2-3n)/2, Sn=(5*1^2-3*1)/2+(5*2^2-3*2)/2+……+(5n^2-3n)/2, =5/2*(1^2+2^2+……+n^2)-3/2*(1+2+……+n) =5/2*(1/6)*(n+1)*(2n+1)-3/2*[n*(n+1)/2] =5/12*(n+1)*(2n+1)-3/4*[n*(n+1)] =1/12*n*(n+1)...

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