Question

请教一道数学化简题，谢谢

化简(3+1)(3^2+1)(3^4+1)(3^8+1)

Answer 1

将式子乘以(3-1),再除以(3-1)
(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)/(3-1)
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)/(3-1)
=(3^4-1)(3^4+1)(3^8+1)/(3-1)
=(3^8-1)(3^8+1)/(3-1)
=(3^16-1)/(3-1)
=(3^16-1)/2

Answer 2

（3+1)(3^2+1)(3^4+1)(3^8+1)
=1/2*(3-1)*(3+1)(3^2+1)(3^4+1)(3^8+1)
=1/2(3^2-1)(3^2+1)(3^4+1)(3^8+1)
=1/2*(3^4-1)(3^4+1)(3^8+1)
=1*2(3^8-1)(3^8+1)
=1/2*(3^16-1)

希望你能找到规律