Question

初一数学：(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)+1=？

百度上不少这个观点
(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=(2-1)[(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)]
=[(2-1)(2+1)](2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=(2^2-1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=[(2^2-1)(2^2+1)](2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=(2^4-1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
...
=(2^2n -1)(2^2n+1)
=2^(2n+2)-1
但是，(2^4-1)(2^4+1)应该是等于(2^4）^2-1吧也就是(2^8）-1啊，那后面的2^6+1怎么办？
求真解

Answer 1

此题有误，应该是
(2+1)(2^2^1+1)(2^2^2+1)(2^2^3+1).......(2^2^n+1)=2^2^(n+1)-1
而，(2^2n+1)(2^2n-1)=2^4n-1

追问

但是不少地方还是拿出来误人子弟，也有不少这样的解答。

Answer 2

(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=(2-1)[(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)]
=[(2-1)(2+1)](2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=(2^2-1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=[(2^2-1)(2^2+1)](2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
=(2^4-1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)
...
=(2^2n -1)(2^2n+1)
=2^(2n+2)-1
没有错

追问

但是，(2^4-1)(2^4+1)应该是等于(2^4）^2-1吧也就是(2^8）-1啊

Answer 3

(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)+1
=（2-1）(2+1)(2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)+1
=（2的平方-1）（2^2+1)(2^4+1)(2^6+1)(2^8+1)……(2^2n+1)+1
=（2的4次方-1）（2^4+1)(2^6+1)(2^8+1)……(2^2n+1)+1
=2的4n次方-1+1
=2的4n次方