(3+1)(3+1)(3+1)(3+1)(3+1)(3+1)

(3+1)(3^2+1)(3^4+1)(3^8+1)…(3^32+1)

解：只需在原式乘上(3-1)/2，然后不断的形成平方差。

原式
=(3-1)/2×(3+1)(3^2+1)(3^4+1)(3^8+1)…(3^32+1)
=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)…(3^32+1)/2
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...(3^32+1)/2
=(3^4-1)(3^4+1)(3^8+1)...(3^32+1)/2
=(3^8-1)(3^8+1)...(3^32+1)/2
=(3^16-1)...(3^32+1)/2
=(3^32-1)(3^32+1)/2
=(3^64-1)/2

(3+1)(3+1)(3+1)(3+1)(3+1)(3+1)=4^6=4096

=4*4*4*4*4*4 =4096

4^6=4096
3^6+6*3^5+15*3^4+20*3^3+15*3^2+6*3+1=4096

(3+1)(3+1)(3+1)(3+1)(3+1)(3+1)
=4*4*4*4*4*4
=4096

有疑问吗?
直接算呗