(2-1)(2+1)(2*2+1)(2*2-1)(2*4+1)........(2*32+1)求详解

来源：百度知道 编辑：UC知道 时间：2024/06/24 13:50:34
要详细过程，越详越好。谢谢

(2-1)(2+1)(2*2+1)(2*2-1)(2*4+1)........(2*32+1)
=(2*2-1)(2*2+1)(2*2-1)(2*4+1)........(2*32+1)
=(2*4-1)(2*4+1)........(2*32+1)
=......
=2^64-1

(2+1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)+1
=[(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)/(2-1)]+1
=[(2^2-1)(2^2+1)(2^4+1)(2^8+1)......(2^32+1)/(2-1)]+1
=[(2^4-1)(2^4+1)(2^8+1)......(2^32+1)/(2-1)]+1
=(2^64-1)/(2-1)+1
=2^64-1+1
=2^64
所以.(2-1)(2+1)(2*2+1)(2*2-1)(2*4+1)........(2*32+1)=2^64-1

从前2项开始用平方差公式：(a-b)(a+b)=a^2-b^2
原式=(2^2-1)(2^2+1)(2^2-1)(2^4+1)……(2^32+1)
=(2^4-1)(2^4+1)……(2^32+1)
=……
=2^64-1

2^64-1