230问答网 > (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)怎么算啊

(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)怎么算啊

急用！！！！

2024-12-21 19:34:23

推荐回答（3个）

回答1：

乘以（3-1）/2变成：
（3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
类推：最后是3^32-1...

回答2：

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

回答3：

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