计算(3+1)(3^2+1)(3^4+1)(3^8+1)-3^16=?

(3+1)(3^2+1)(3^4+1)(3^8+1)-3^16=?
2024-12-29 13:40:55
推荐回答(2个)
回答1:

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

回答2:

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