2(3^2+1)(3^4+1)(3^6+1)(3^8+1)要简算

问题描述:

2(3^2+1)(3^4+1)(3^6+1)(3^8+1)要简算

2=(3^2-1)/4
令原式=A

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

=(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)(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