(1+2)(1+2的平方)(1+2的4次方)(1+2的8次方)=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/30 00:00:56

(1+2)(1+2的平方)(1+2的4次方)(1+2的8次方)=?
(1+2)(1+2的平方)(1+2的4次方)(1+2的8次方)=?

(1+2)(1+2的平方)(1+2的4次方)(1+2的8次方)=?
(1+2)(1+2^2)(1+2^4)(1+2^8)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)
=(2^4-1)(2^4+1)(2^8+1)
=(2^8-1)(2^8+1)
=2^16-1
提示:前面加2-1,然后反复用平方差公式.

(1+2)(1+2^2)(1+2^4)(1+2^8)
=(1-2)(1+2)(1+2^2)(1+2^4)(1+2^8)/(1-2)
=(1-2^2)(1+2^2)(1+2^4)(1+2^8)/(1-2)
=(1-2^4)(1+2^4)(1+2^8)/(1-2)
=(1-2^8)(1+2^8)/(1-2)
=(1-2^16)/(1-2)
=2^16-1

只需要增加一项(2-1):
(1+2)(1+2^2)(1+2^4)(1+2^8)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)
=(2^4-1)(2^4+1)(2^8+1)
=(2^8-1)(2^8+1)
=2^16-1.

现在前面×(1-2)即可 即配成平方差公式

(1+2)(1+2^2)(1+2^4)(1+2^8)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1)
=(2^4-1)(2^4+1)(2^8+1)
=(2^8-1)(2^8+1)
=2^16-1
配成平方差公式