..{1+[1/(2^1)]}*{1+[1/(2^2)]}*{1+[1/(2^4)]}*{1+[1/(2^8)]}+[1

有没有学过平方差公式,即(a+b)*(a-b)=a^2 - b^2.这题利用这个可以轻松完成,如下：
原式={[1-(1/2^1)]/[1-(1/2^1)]}*{1+[1/(2^1)]}*{1+[1/(2^2)]}*{1+[1/(2^4)]}*{1+[1/(2^8)]}+[1/(2^15)]
={1+[1/(2^1)]}*{1-[1/(2^1)]}*{1+[1/(2^2)]}*{1+[1/(2^4)]}*{1+[1/(2^8)]}/{1-[1/(2^1)]}+[1/(2^15)]
(这里a=1,b=1/(2^1),利用平方差公式a^2 - b^2=1-[1/(2^2)])
={1-[1/(2^2)]}*{1+[1/(2^2)]}*{1+[1/(2^4)]}*{1+[1/(2^8)]}/{1-[1/(2^1)]}+[1/(2^15)]
(一下不断使用平方差公式)
={1-[1/(2^4)]}*{1+[1/(2^4)]}*{1+[1/(2^8)]}/{1-[1/(2^1)]}+[1/(2^15)]
={1-[1/(2^8)]}*{1+[1/(2^8)]}/{1-[1/(2^1)]}+[1/(2^15)]
={1-[1/(2^16)]}/{1-[1/(2^1)]}+[1/(2^15)]
=2 * {1-[1/(2^16)]} + [1/(2^15)]
=2 - [2* 1/(2^16)] + [1/(2^15)]
=2 - [1/(2^15)] + [1/(2^15)]
=2
不用随便说什么难啊,超难的,动动脑筋就可以想出来了!

能用图片发上来吗？

{1+[1/(2^1)]}*{1+[1/(2^2)]}*{1+[1/(2^4)]}*{1+[1/(2^8)]}+[1/(2^15)]
={(1-1/2^1)(1+1/2^1)(1+1/2^2)(1+1/2^4)(1+1/2^8)}+1/2^15
=2(1-1/2^2)(1+1/2^2)(1+1/2^4)(1+1/2^8)}+1/2^15
=2(1-1/2^4)(1+1/2^4)(1+1/2^8)}+1/2^15
=2(1-1/2^8)(1+1/2^8)}+1/2^15
=2(1-1/2^16)+1/2^15
=2-1/2^15+1/2^15
=2

{1+[1/(2^1)]}*{1+[1/(2^2)]}*{1+[1/(2^4)]}*{1+[1/(2^8)]}+[1/(2^15)]
=3/2*5/4*17/16*257/256+2^15
=2