2(3^2+1）(3^4+1)(3^6+1)(3^8+1)要简算

来源：学生作业帮助网编辑：作业帮时间：2024/04/29 05:00:51

2(3^2+1）(3^4+1)(3^6+1)(3^8+1)要简算
2(3^2+1）(3^4+1)(3^6+1)(3^8+1)要简算

2(3^2+1）(3^4+1)(3^6+1)(3^8+1)要简算
2(3^2+1）(3^4+1)(3^6+1)(3^8+1)=2(3^2+1）(3^4+1)(3^6+1)(3^8+1)*(3^2-1）/(3^2-1)
=2*(3^4-1)(3^4+1)(3^6+1)(3^8+1)/(3^2-1)
=2*(3^8-1)(3^6+1)(3^8+1)/(3^2-1)
=2*(3^16-1)(3^6+1)/(3^2-1)
=[3^22+3^16-3^6-1]/4

=(3-1)(3+1)(3^2+1）(3^4+1)(3^6+1)(3^8+1)/(3+1)
=(3^16-1)(3^6+1)/4

2＝(3^2-1)/4
令原式＝A
则
4A=(3^2-1)(3^2+1)(3^4+1)(3^6+1)(3^8+1)
=(3^4-1)(3^4+1)(3^6+1)(3^8+1)
=(3^8-1)(3^8+1)(3^6+1)
=(3^16-1)(3^6+1)
即原式＝A=(3^16-1)(3^6+1)/4