Ta có: 3 + 1 = (3^2 - 1)/(3 - 1)
3^2 + 1 = (3^4 - 1)/(3^2 - 1)
3^4 + 1 = (3^8 - 1)/(3^4 - 1)
3^8 + 1 = (3^16 - 1)/(3^8 - 1)
3^16 + 1 = (3^32 - 1)/(3^16 - 1)
3^32 + 1 = (3^64 - 1)/(3^32 - 1)
(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^16 + 1)(3^32 + 1)
=(3^2 - 1)/(3 - 1).(3^4 - 1)/(3^2 - 1).(3^8 - 1)/(3^4 - 1).(3^32 - 1)/(3^16 - 1).(3^64 - 1)/(3^32 - 1)
=(3^64 - 1)/(3 - 1)
=(3^64 - 1)/2
Đặt biểu thức đó là A
(3-1) A= (3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1) (3^32+1)
2 A= (3^2-1)(3^2+1)(3^4+1)..............................................
2A = (3^4-1)(3^4+1)(3^8+1) ............................
2A= (3^8-1)(3^8+1)(3^16+1) .............
2A = (3^16-10(3^16+1)(3^32+1)
2A = (3^32-1)(3^32+1)
2A= 3^64-1
A= (3^64-1) / 2
#)Giải :
\(=\frac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{8}\)
\(=\frac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{8}\)
\(=\frac{\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{8}\)
\(=\frac{\left(3^{32}-1\right)\left(3^{32}+1\right)}{8}\)
\(=\frac{3^{64}-1}{8}\)