Question

(2+1)×(2平方+1)×(2的四次方+1)×······×(2的32次方+1)

Answer 1

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

^表示指数。
提示：就是反复利用平方差公式。

Answer 2

(2 + 1)(2^2 + 1)(2^4 + 1) ... ( 2^32 + 1)
= ( 2 - 1 ) (2 + 1)(2^2 + 1)(2^4 + 1) ... ( 2^32 + 1)
= (2^2 - 1)(2^2 + 1)(2^4 + 1) ... ( 2^32 + 1)
= (2^4 - 1)(2^4 + 1) ... ( 2^32 + 1)

由这个规律继续下去：

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