Question

计算:(2+1)(2^2+1)(2^4+1)……(2^32+1)+1

Answer 1

解：因为(2-1)=1
所以可以给原式乘上(2-1)，原式的值不变
原式=(2-1)(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)+1
=(2²-1)(2²+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

追问

那加上（3-2）不行吗？

追答

加(2-1)是为了能与(2+1)形成(a-b)(a+b)的形式，所以只能加(2-1)

Answer 2

(2+1)(2^2+1)(2^4+1)……(2^32+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)……(2^32+1)+1
=(2^2-1)(2^2+1)(2^4+1)……(2^32+1)+1
=(2^4-1)(2^4+1)……(2^32+1)+1
=…
=(2^32-1)(2^32+1)+1
=2^64-1+1
=2^64

Answer 3

(2+1)(2^2+1)(2^4+1)……(2^32+1)+1
=(2-1)(2+1)(2^2+1)(2^4+1)……(2^32+1)+1
=(2^2-1)(2^2+1)(2^4+1)……(2^32+1)+1
=……
=2^64