Question

数学题：（3+1）（3的平方+1）（3的四次方+1）（3的八次方+1）（3的十六次方+1）（3的三十二次方+1）=？

急，快！！！！！

Answer 1

(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)=？

解：原式通过乘上1/2×(3-1)，可以不断的形成平方差。
原式
=1/2×(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=1/2×(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=1/2×(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=1/2×(3^8-1)(3^8+1)(3^16+1)(3^32+1)
=1/2×(3^16-1)(3^16+1)(3^32+1)
=1/2×(3^32-1)(3^32+1)
=1/2×(3^64-1)
=(3^64-1)/2

Answer 2

原式=（3-1)(3+1)(3^2+1)....(3^32+1)/(3-1)

=(3^2-1)(3^2+1)....(3^32+1)/2

=(3^4-1).....(3^32+1)/2

=....

=(3^32-1)(3^32+1)/2

=(3^64-1)/2