问题描述: 计算2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1 1个回答 分类:数学 2014-10-04 问题解答: 我来补答 反复运用公式(a-b)(a+b) = a^2-b^2.2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1= (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1= (3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1= (3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)+1= (3^8-1)(3^8+1)(3^16+1)(3^32+1)+1= (3^16-1)(3^16+1)(3^32+1)+1= (3^32-1)(3^32+1)+1= 3^64-1+1= 3^64. 展开全文阅读