(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?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/08 12:50:19
(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?
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
(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)
.
.
.
=(2^64-1)(2^64+1)
=2^128-1
乘以(2-1)
原式=2^128-1
1/2,1/4,
(1-1/2^2)(1-1/3^2)(1-1/4^2).(1-1/2009^2),
(2+1)({2}^{2}+1)({2}^{4}+1)({2}^{8}+1).({2}^{64}+1)+1
巧算((2^1+1)(2^2+1)(2^4+1)(2^8+1)+1)/2^15
计算(2+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)+1
[(1+2^-(1/32)]*[(1+2^-(1/16)]*[(1+2^-(1/8)]*[(1+2^-(1/4)]*[(1+2^-(1/2)]
(1-1/2^2)*(1-1/3^2)*(1-1/4^2)*.*(1-1/2002^2)*(1-1/2003^2)
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)(1+1/2^16),
1,2,4,4,1,( )
(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^16
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15
计算题:(1+1/2) (1+1/2^2) (1+1/2^4) (1+1/2^8) +1/2^15
(1+1/2)(1+1/2^2)(1+1/2^4)(1+1/2^8)+1/2^15=?
1、1/2、1/4、1/7
计算:(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.计算:(2+1)(2^2+1)(2^4+1)(2^8+1)-2^16.
求和:sn=1/2^2-1+1/4^2-1+.1/(2n)^2-1
(2^2+1)*(2^4+1)*(2^8+1)*(2^16+1)*(2^32+1)