(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)怎么算
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/25 10:48:57
(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)怎么算
![(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)怎么算](/uploads/image/z/1283968-64-8.jpg?t=%283%2B1%29%283%5E2%2B1%EF%BC%89%EF%BC%883%5E4%2B1%29%283%5E8%2B1%29%283%5E16%2B1%29%E6%80%8E%E4%B9%88%E7%AE%97)
乘一个(3-1)/2啊
利用公式(a-b)*(a+b)=a^2-b^2
(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
=...
=(3^16-1)(3^16+1)/2
=(3^32-1)/2
利用公式(a-b)*(a+b)=a^2-b^2
(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
=...
=(3^16-1)(3^16+1)/2
=(3^32-1)/2