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计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
急``!
人气:104 ℃ 时间:2020-04-13 21:44:11
解答
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
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