(3+1)(3^2+1)(3^4+1)(3^8+1)+(3^16+1)=?

过程
2024-12-21 08:07:58
推荐回答(1个)
回答1:

原式=[(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
······以此类推······
=(3^32-1)/2