(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/04 11:14:06
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
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(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
题没有打完
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1
=(2²-1) (2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),/3+1
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),/3+1
=...
=(2^128-1)/3 +1
先乘以(2²-1),再除以3,保证相等
然后反复应用平方差公式
=(2²-1) (2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),/3+1
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),/3+1
=...
=(2^128-1)/3 +1
先乘以(2²-1),再除以3,保证相等
然后反复应用平方差公式
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256
1/2-1/4-1/8-16/1-32/1-64/1-128/1-256/1
已知m=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128
设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),试确定M-2
设M=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)(2^128+
求解以下两式的值:1、(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末位数字.
试确定(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)+1的末位数字
计算:(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)
(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1)等于多少?