Question

数学题(2+1)(2的平方+1)(2的4次方+1)...(2的32次方+1)+1怎么算?

Answer 1

1楼的解法是对的，不过他头几步后面一个+1给漏掉了

这里用到一个技巧，就是前面那一堆
(2+1)(2^2+1)(2^4+1)(2^8+1).......(2^32+1)
多乘了一个 1 就是
1*(2+1)(2^2+1)(2^4+1)(2^8+1).......(2^32+1)
由于2-1=1
故：
（2-1）*(2+1)(2^2+1)(2^4+1)(2^8+1).......(2^32+1)+1
=(2^2-1)(2^2+1)(2^4+1)(2^8+1).......(2^32+1)
=(2^4-1)(2^4+1)(2^8+1).......(2^32+1)+1
=(2^8-1)(2^8+1).......(2^32+1)+1
=(2^16-1).......(2^32+1)+1
.......
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64

Answer 2

2^64 吧，应该不会错

Answer 3

(2+1)(2^2+1)(2^4+1)(2^8+1).......(2^32+1)
=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1).......(2^32+1)
=(2^2-1)(2^2+1)(2^4+1)(2^8+1).......(2^32+1)
=(2^4-1)(2^4+1)(2^8+1).......(2^32+1)+1
=(2^8-1)(2^8+1).......(2^32+1)+1
=(2^16-1).......(2^32+1)+1
.......
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64

追问

(2-1)(2+1)是由什么转变而来的？

追答

2-1还是1啊
相当没有 只是要平方差