Question

计算:(1+½)(1+1/2²)(1+2的四次方分之一)(1+2的十六次方分之一) 要过程。。。。急啊

过程，要巧算

Answer 1

(1+½)(1+1/2²)(1+1/2^4)(1+1/2^8)(1+1/2^16)
=2(1-1/2)(1+½)(1+1/2²)(1+1/2^4)(1+1/2^8)(1+1/2^16)
=2(1-1/2²)(1+1/2²)(1+1/2^4)(1+1/2^8)(1+1/2^16)
=2(1-1/2^4)(1+1/2^4)(1+1/2^8)(1+1/2^16)
=2(1-1/2^8)(1+1/2^8)(1+1/2^16)
=2(1-1/2^16)(1+1/2^16)
=2(1-1/2^32)
=2 -1/2^31

^表示指数，2^31表示2的31次方。

Answer 2

(2²+1)(2^4+1)(2^8+1)(2^16+1）
=（2^2-1)(2²+1)(2^4+1)(2^8+1)(2^16+1）/(2^2-1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1）/3
= (2^8-1)(2^8+1)(2^16+1）/3
=(2^16-1)(2^16+1）/3
=(2^32-11)/3
=(2^32-1)/3

追问

(2²+1)(2^4+1)(2^8+1)(2^16+1）题目不是这个