计算：2(3+1)(3^2+1)(3^4+1）.(3^16+1)+1
问题描述：
计算：2(3+1)(3^2+1)(3^4+1）.(3^16+1)+1
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