问题描述： (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)怎么算啊急用! 1个回答 分类：综合 2014-10-28 问题解答： 我来补答 (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)-1 =1/2*(3-1)*(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)-1 =1/2(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)-1 =1/2*(3^4-1)(3^4+1)(3^8+1)(3^16+1)-1 =1*2(3^8-1)(3^8+1)(3^16+1)-1 =1/2*(3^16-1)(3^16+1)-1 =1/2(3^32-1)-1 =1/2(3^32+1) =3^32/2+1/2 展开全文阅读
