7/8/{(1/2-1/4)*3/16}

此题初见可知,分母的表现形式为乘积且存在等差关系,分子为1,求几个数字之和
1.我们当然不应该硬算,题目有一定规律,从等差入手,把题目视为求数列之和,找出“通项”（每个乘号左边1,4,7...求出通项3n-2,每个乘号右边4,7,10.求出通项3n+1,）所以通项可视为1/(3n
-2)*(3n+1)
2求和,从以上通项可知1/(3n-2)*(3n+1)=[1/(3n-2)-1/(3n+1)]*1/3,然后就能把原式进行拆分,消去同项,进行简易求和,最后是（1-1/22)*1/3=7/22
楼主看在数学题难打的情况上给个采纳吧.

答案:n/(n+1)
解:(用裂项相消法)
s=1/2+1/(2+4)+1/(2+4+6)+1/(2+4+6+8)+...+1/(2+4+6+8+...+2n)
=1/2+1/6+1/12+1/20+…+1/(2+4+6+8+…+2n)
=1/(1*2)+1/(2*3)+1/(3*4)+1/(4*5)+…+1/[n*(n+1)]
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)

1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+1/(4*5*6)+.+1/(24*25*26)
=1/2*(1/1*2-1/2*3)+1/2*(1/2*3-1/3*4)+……+1/2*(1/24*25-1/25*26)
=1/2*(1/1*2-1/2*3+1/2*3-1/3*4+……+1/24*25-1/25*26)
=1/2*(1/1*2-1/25*26)
=1/2*(1/2-1/650)
=81/325

1.960602933

1/(1*2*3*4)=1/3(1/1*2*3-1/2*3*4)
1/(2*3*4*5)=1/3(1/2*3*4-1/3*4*5)
..
...
.
1/(1*2*3*4)+1/(2*3*4*5)+1/(3*4*5*6).+1/(17*18*19*20)
=1/3(1/1*2*3-1/18*19*20)
=1/3*(3*19*20-1)/(18*19*20)
=1139/20520

提示一下,1+2+3+…+n=n(n+1)/2

负55分之六十二

(5/12-1/4)/9/4
=1/6÷9/4
=2/27
2/3-(2/3-1/2)
=2/3-2/3+1/2
=1/2
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1/(2*3*4)=1/2*3-(1/3*4)
1/(3*4*5)=1/3*4-(1/4*5)
以此类推

1.(x^2+2x-2)(x^2+2x+4)+9
=(x^2+2x)^2+2(x^2+2x)+1
=(x^2+2x+1)^2
=(x+1)^4
2.(a^2-1)^2+30(1-a^2)+225
=(a^2-1)^2-30(a^2-1)+225
=(a^2-1-15)^2
=[(a+1)(a-1)]^2
3.(x-y)^2+4(x^2-y^2)+4(x+y)^2
=(x-y)^2+4(x-y)(x+y)+4(x+y)^2
=(x-y+2x+2y)^2
=(3x+y)^2