计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)急``!
问题描述：
计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
急``!
答
思路：在原式乘上（2-1）,不断的产生平方差,可以巧解.
(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
原式=(2-1)(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
=(2^16-1)(2^16+1)(2^32+1)(2^64+1)
=(2^32-1)(2^32+1)(2^64+1)
=(2^64-1)(2^64+1)
=2^128-1