化简:(1+2^1⼀32)(1+2^1⼀16)(1+2^1⼀8)(1+2^1⼀4)(1+2^1⼀2)

2024-12-17 20:07:49
推荐回答(2个)
回答1:

设S=(1+2^-1/32)(1+2^-1/16)(1+2^-1/8)(1+2^-1/4)(1+2^-1/2)
(1-2^-1/32)*S=(1-2^-1/32)(1+2^-1/32)(1+2^-1/16)(1+2^-1/8)(1+2^-1/4)(1+2^-1/2)
[由于(1-2^-1/32)(1+2^-1/32)=1-2^-16]
(1-2^-1/32)*S=(1-2^-16)(1+2^-1/16)(1+2^-1/8)(1+2^-1/4)(1+2^-1/2)[由于(1-2^-1/16)(1+2^-1/16)=1-2^-1/8]
(1-2^-1/32)*S=(1-2^-1/8)(1+2^-1/8)(1+2^-1/4)(1+2^-1/2)=(1-2^-1/4)(1+2^-1/4)(1+2^-1/2)=(1-2^-1/2)(1+2^-1/2)=1-2^-1=1/2
所以S=(1/2)/[1-2^(-1/32)]

回答2:

原式左乘一个[1-2^(1/32)]/[1-2^(1/32)]得
[1-2^(1/32)]*[1+2^(1/32)]*[1+2^(1/16)]*[1+2^(1/8)]*[1+2^(1/4)]*[1+2^(1/2)]/[1-2^(1/32)]
=[1-2^(1/16)]*[1+2^(1/16)]*[1+2^(1/8)]*[1+2^(1/4)]*[1+2^(1/2)]/[1-2^(1/32)]
=[1-2^(1/8)]*[1+2^(1/8)]*[1+2^(1/4)]*[1+2^(1/2)]/[1-2^(1/32)]
=[1-2^(1/4)]*[1+2^(1/4)]*[1+2^(1/2)]/[1-2^(1/32)]
=[1-2^(1/2)]*[1+2^(1/2)]/[1-2^(1/32)]
=(1-2)/[1-2^(1/32)]
=1/[2^(1/32)-1]