2(3^2+1）(3^4+1)(3^6+1)(3^8+1)要简算_数学解答

2(3^2+1）(3^4+1)(3^6+1)(3^8+1)要简算

人气：324 ℃　时间：2020-07-27 11:49:22

解答

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