搜搜考试网(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：数学作业 时间：2024/06/23 18:09:58
(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^1-1)(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^16-1)(2^16+1)(2^32+1)+1
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64
再问： 请问能讲解一下吗？
再答： 2^1-1=1 运用平方差公式
=(2^1-1)(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^8-1)(2^8+1)(2^16+1)(2^32+1)+1
=(2^16-1)(2^16+1)(2^32+1)+1
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64
再问： 请问能讲解一下吗？
再答： 2^1-1=1 运用平方差公式
1/2+1/4+1/8+1/16+1/32+.
(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1计算方法是什么?
(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1=
计算(1+2)(1+2^2;)(1+2^4)(1+2^8)(1+2^16)(1+2^32)
(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)(2^64+1),
计算:(1+1/2^32)(1+1/2^16)(1+1/6^8)(1+1/2^4)(1+1/2^2)(1+1/2) .
计算3(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1的值
计算:3*(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)+1
计算(1+2-1/32)(1+2-1/16)(1+2-1/8)(1+2-1/4)(1+2-1/2) 注：每个括号内的-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