作业帮计算:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/（2^32）-1=?
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/04/28 23:49:03
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/（2^32）-1=?
:(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^8-1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^16-1)(2^16+1)/[（2^32）-1]
=(2^32-1)/[（2^32）-1]
=1
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^8-1)(2^8+1)(2^16+1)/[（2^32）-1]
=(2^16-1)(2^16+1)/[（2^32）-1]
=(2^32-1)/[（2^32）-1]
=1