计算：2(3+1)(3^2+1)(3^4+1）.(3^16+1)+1_数学解答

计算：2(3+1)(3^2+1)(3^4+1）.(3^16+1)+1

人气：207 ℃　时间：2020-03-29 22:34:02

解答

2(3+1)(3^2+1)(3^4+1）.(3^16+1)+1
=(3-1)(3+1)(3^2+1)(3^4+1）.(3^16+1)+1
=(3^2-1)(3^2+1)(3^4+1）.(3^16+1)+1
=(3^4-1)(3^4+1）.(3^16+1)+1
反复用平方差
=3^32-1+1
=3^32