初一(1+1/2)(1+1/4)(1+1/16)(1+1/256)(1-1/2)*(1+1/2)*(1+1/4)*(1+1/16))*(1+1/256)/(1-1/2)=(1-1/256*1/256)/(1-1/2)=2-2/256*256

问题描述:

初一(1+1/2)(1+1/4)(1+1/16)(1+1/256)
(1-1/2)*(1+1/2)*(1+1/4)*(1+1/16))*(1+1/256)/(1-1/2)
=(1-1/256*1/256)/(1-1/2)
=2-2/256*256

(1-1/2)*(1+1/2)=1-1/4;
(1-1/4)*(1+1/4)=1-1/16;
(1-1/16)*(1-1/16)=1-1/256;
(1-1/256)*(1+1/256)=1-1/256*256;
1-1/2=1/2;
推得

初一就有点超纲了 用到平方差公式(a+b)(a-b)=a^2-b^2所以
(1-1/2)*(1+1/2)=(1-1/4)
(1-1/4)*(1+1/4)=1-1/16
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所以要化简的式子的分子变成了(1-1/256)(1+1/256)=1-1/256*1/256

不停地用平方差公式,前两个括号乘起来是1-1/4,和后面一个式子合起来再用平方差公式。

(1-1/2)*(1+1/2)=(1-1/4)
以此类推就可以得到答案了

平方差公式.(1-1/2)(1+1/2)=1^2-(1/2)^2=1-1/4(1-1/4)(1+1/4)=1^2-(1/4)^2=1-1/16(1-1/16)(1+1/16)=1^2-(1/16)^2=1-1/256(1-1/256)(1+1/256)=1^2-(1/256)^2(1+1/2)(1+1/4)(1+1/16)(1+1/256)=(1-1/2)*(1+1/2)*(1...

(1-1/2)(1+1/2)=(1-1/4).则(1-1/2)(1+1/2)(1+1/4)=(1-1/4)(1+1/4)=(1-1/16).以此类推最后得到(1-1/256*1/256)

用了平方差
乘以1-1/2,在除以1-1/2
就得到(1-1/2)*(1+1/2)*(1+1/4)*(1+1/16))*(1+1/256)/(1-1/2)
然后(1-1/2)*(1+1/2)=1²-(1/2)²=1-1/4
所以(1-1/2)*(1+1/2)*(1+1/4)
=(1-1/4)*(1+1/4)
=1-1/16
反复用平方差
且分母1-1/2=1/2
就是(1-1/256²)/(1/2)
=2
(1-1/256²)
=2-2/256²